CodeVs刷题攻略之Gold

CodeVs刷题攻略之Gold

2018.01.04 By gwj233

0x01贪心

  1. 均分纸牌

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    #include<iostream>
    using namespace std;
    const int maxn = 110;
    int a[maxn], sum, ans;
    int main(){
    int n; cin>>n;
    for(int i = 1; i <= n; i++){
    cin>>a[i]; sum += a[i];
    }
    sum /= n;
    for(int i = 1; i <= n; i++){
    if(a[i] != sum){
    ans++;
    a[i+1] -= sum-a[i];
    }
    }
    cout<<ans<<'\n';
    return 0;
    }
  2. 线段覆盖

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    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int maxn = 110;
    int a[maxn], b[maxn], c[maxn];
    bool cmp(int x, int y){
    return b[x]==b[y] ? a[x]<a[y] : b[x]<b[y];
    }
    int main(){
    int n; cin>>n;
    for(int i = 1; i <= n; i++){
    int x, y; cin>>x>>y; //输入有坑,可能先大再小
    a[i]=min(x,y); b[i]=max(x,y); c[i]=i;
    }
    sort(c+1,c+n+1,cmp);
    int t = -999, ans = 0;
    for(int i = 1; i <= n; i++){
    if(t <= a[c[i]]){
    t = b[c[i]]; ans++;
    }
    }
    cout<<ans<<"\n";
    return 0;
    }

0x02高精度入门

  1. 高精度练习之减法

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    #include<iostream>
    #include<algorithm>
    #include<string>
    using namespace std;
    const int maxn = 510;
    int a[maxn], b[maxn], c[maxn];
    int main(){
    string s1, s2;
    cin>>s1>>s2;
    if(s1.size()<s2.size()||(s1.size()==s2.size()&&s1<s2)){ cout<<"-"; swap(s1,s2); }
    a[0] = s1.size();
    b[0] = s2.size();
    for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
    for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
    c[0] = max(a[0], b[0]);
    for(int i = 1; i <= c[0]; i++){
    if(a[i]<b[i]){ a[i+1]--; a[i]+=10; }
    c[i] = a[i]-b[i];
    }
    while(c[0]>1 && c[c[0]]==0)c[0]--;
    for(int i = c[0]; i >= 1; i--)cout<<c[i];
    return 0;
    }
  2. 高精度练习之加法

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    #include<iostream>
    #include<algorithm>
    #include<string>
    using namespace std;
    const int maxn = 510;
    int a[maxn], b[maxn], c[maxn];
    int main(){
    string s1, s2;
    cin>>s1>>s2;
    a[0] = s1.size();
    b[0] = s2.size();
    for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
    for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
    c[0] = max(a[0], b[0])+1;
    for(int i = 1; i <= c[0]; i++){
    c[i] += a[i]+b[i];
    if(c[i]>=10){ c[i+1]++; c[i]%=10; }
    }
    while(c[0]>1 && c[c[0]]==0)c[0]--;
    for(int i = c[0]; i >= 1; i--)cout<<c[i];
    return 0;
    }
  3. 高精度练习之乘法

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    #include<iostream>
    #include<algorithm>
    #include<string>
    using namespace std;
    const int maxn = 510;
    int a[maxn], b[maxn], c[maxn];
    int main(){
    string s1, s2;
    cin>>s1>>s2;
    a[0] = s1.size();
    b[0] = s2.size();
    for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
    for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
    c[0] = a[0]+b[0];
    for(int i = 1; i <= a[0]; i++){
    for(int j =1; j <= b[0]; j++){
    c[i+j-1] += a[i]*b[j];
    if(c[i+j-1]>=10){
    c[i+j] += c[i+j-1]/10;
    c[i+j-1] %= 10;
    }
    }
    }
    while(c[0]>1 && c[c[0]]==0)c[0]--;
    for(int i = c[0]; i >= 1; i--)cout<<c[i];
    return 0;
    }

0x03背包型动态规划

  1. 装箱问题

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    //数据太水,搜索就好
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, m, a[110];
    int search(int i, int r){
    if(i == n)return r;
    if(r < a[i])return search(i+1, r);
    return min(search(i+1,r), search(i+1,r-a[i]));
    }
    int main(){
    cin>>m>>n;
    for(int i = 0; i < n; i++)cin>>a[i];
    cout<<search(0,m);
    return 0;
    }
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    //f[i][j],表示在选第i个物品时, 大小为j的体积能否取到
    #include<iostream>
    using namespace std;
    const int maxn = 20010;
    int n, m, a[35], f[35][maxn];
    int main(){
    cin>>m>>n;
    for(int i = 1; i <= n; i++)cin>>a[i];
    f[0][0] = 1;
    for(int i = 1; i <= n; i++)
    for(int j = 0; j <= m; j++)
    if(f[i-1][j]){
    f[i][j] = f[i-1][j];
    if(j+a[i]<=m)f[i][j+a[i]] = 1;
    }
    for(int i = m; i >= 0; i--)
    if(f[n][i]){ cout<<m-i<<"\n"; return 0;}
    return 0;
    }
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    //f[i]表示大小为i的体积能否取到
    #include<iostream>
    using namespace std;
    const int maxn = 20010;
    int n, m, a[maxn], f[maxn];
    int main(){
    cin>>m>>n;
    for(int i = 1; i <= n; i++)cin>>a[i];
    f[0] = 1;
    for(int i = 1; i <= n; i++)
    for(int j = m; j >= 0; j--)
    if(f[j] && j+a[i]<=m)
    f[j+a[i]] = 1;
    for(int i = m; i >= 0; i--)
    if(f[i]){ cout<<m-i<<"\n"; return 0;}
    return 0;
    }
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    //01背包, f[i]表示体积为i时得到的最大价值
    #include<iostream>
    using namespace std;
    const int maxn = 20010;
    int n, m, w[35], v[35], f[maxn];
    int main(){
    cin>>m>>n;
    for(int i = 1; i <= n; i++){
    cin>>w[i]; v[i] = w[i];
    }
    for(int i = 1; i <= n; i++)
    for(int j = m; j >= w[i]; j--)
    f[j] = max(f[j],f[j-w[i]]+v[i]);
    cout<<m-f[m]<<"\n";
    return 0;
    }
  2. 乌龟棋

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    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, m, a[350], num[5], f[50][50][50][50];
    int main(){
    cin>>n>>m;
    for(int i = 1; i <= n; i++)cin>>a[i];
    for(int i = 1; i <= m; i++){
    int x; cin>>x; num[x]++;
    }
    for(int i = 0; i <= num[1]; i++){
    for(int j = 0; j <= num[2]; j++){
    for(int k = 0; k <= num[3]; k++){
    for(int l = 0; l <= num[4]; l++){
    int t = 0;
    if(i)t = max(t, f[i-1][j][k][l]);
    if(j)t = max(t, f[i][j-1][k][l]);
    if(k)t = max(t, f[i][j][k-1][l]);
    if(l)t = max(t, f[i][j][k][l-1]);
    f[i][j][k][l] = t+a[i+j*2+k*3+l*4+1];
    }
    }
    }
    }
    cout<<f[num[1]][num[2]][num[3]][num[4]]<<"\n";
    return 0;
    }

0x04序列型动态规划

  1. 拦截导弹

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    //LDS的个数等于LIS的长度
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a[30], f[30], ans;
    int main(){
    while(cin>>a[n])n++;
    for(int i = 0; i < n; i++)f[i] = 1;
    for(int i = 0; i < n; i++)
    for(int j = 0; j < i; j++)
    if(a[j]>=a[i])f[i] = max(f[i], f[j]+1);
    for(int i = 0; i < n; i++)ans = max(ans, f[i]);
    cout<<ans<<"\n";
    for(int i = 0; i < n; i++)f[i] = 1;
    for(int i = 0; i < n; i++)
    for(int j = 0; j < i; j++)
    if(a[j]<a[i])f[i] = max(f[i], f[j]+1);
    ans = 0;
    for(int i = 0; i < n; i++)ans = max(ans, f[i]);
    cout<<ans<<"\n";
    return 0;
    }
  2. 最长严格上升子序列

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    //f[i]:到i为止的LIS的长度。
    //f[i]=max{1,f[j]+1|j<i&&aj<ai}
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a[510], f[510], ans;
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)cin>>a[i];
    for(int i = 1; i <= n; i++){
    int t = 0;
    for(int j = 1; j < i; j++)
    if(a[j]<a[i])t = max(t, f[j]);
    f[i] = t+1;
    ans = max(ans, f[i]);
    }
    cout<<ans<<"\n";
    return 0;
    }
  3. 线段覆盖 2

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    //f[i]:到第i条线段为止能获得的最大价值
    //f[i]=max{s[i].c,f[j]+s[i].c|j<i&&s[j].b<=s[i].a}
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int maxn = 1010;
    struct seg{ int a, b, c; }s[maxn];
    bool cmp(seg a, seg b){ return a.a<b.a; }
    int n, f[maxn], ans;
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)
    cin>>s[i].a>>s[i].b>>s[i].c;
    sort(s+1,s+n+1,cmp);
    for(int i = 1; i <= n; i++){
    f[i] = s[i].c;
    for(int j = 1; j < i; j++)
    if(s[j].b<=s[i].a)f[i] = max(f[i], f[j]+s[i].c);
    ans = max(ans, f[i]);
    }
    cout<<ans<<"\n";
    return 0;
    }

0x05区间型动态规划

  1. 石子归并

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    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int maxn = 1010, inf=0xffffff;
    int w[maxn], f[maxn][maxn];
    int main(){
    int n; cin>>n;
    for(int i = 1; i <= n; i++)cin>>w[i];
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n; j++)
    f[i][j] = i==j ? 0 : inf;
    for(int i = 1; i <= n; i++)w[i] += w[i-1];
    //转移顺序长度从小到大覆盖即可
    for(int i = n; i >= 1; i--)//枚举起点
    for(int j = i; j <= n; j++)//枚举终点
    for(int k = i; k <= j; k++)//枚举断点
    f[i][j] = min(f[i][j],f[i][k]+f[k+1][j]+w[j]-w[i-1]);
    cout<<f[1][n]<<"\n";
    return 0;
    }
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    //解释一下, DP题写几个版本主要看心情。。。
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int maxn = 1010, inf=0xffffff;
    int w[maxn], f[maxn][maxn];
    int dp(int i, int j){
    if(f[i][j] != inf)return f[i][j];
    for(int k = i; k <= j; k++)
    f[i][j] = min(f[i][j], dp(i,k)+dp(k+1,j)+w[j]-w[i-1]);
    return f[i][j];
    }
    int main(){
    int n; cin>>n;
    for(int i = 1; i <= n; i++)cin>>w[i];
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n; j++)
    f[i][j] = i==j ? 0 : inf;
    for(int i = 1; i <= n; i++)w[i] += w[i-1];
    cout<<dp(1,n)<<"\n";
    return 0;
    }
  2. 能量项链

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    //环形DP, 复制一份, 拆环成链,然后做区间DP(即对每条链做一遍区间DP,取最大值。)
    //解释一下, 注释详细度主要看心情。。。
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int maxn = 1010;
    int w[maxn], f[maxn][maxn];
    int dp(int i, int j){
    if(f[i][j])return f[i][j];
    //注意边界, [i~i,i+1~j] and [i~j-1,j~j];
    for(int k = i; k <= j-1; k++)
    //通过k合并后, 两颗珠子分别为[i,k],[k+1,j]其中计算能量时headi*taili*headj;
    f[i][j] = max(f[i][j], dp(i,k)+dp(k+1,j)+w[i]*w[k+1]*w[j+1]);
    return f[i][j];
    }
    int main(){
    int n; cin>>n;
    for(int i = 1; i <= n; i++){ cin>>w[i]; w[i+n]=w[i]; }
    dp(1, 2*n);
    int ans = 0;
    for(int i = 1; i <= n; i++)//扫描每条长度为n的链, 最大值为答案。
    ans = max(ans, f[i][i+n-1]);
    cout<<ans<<'\n';
    return 0;
    }
  3. 矩阵取数游戏

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    //每行独立区间DP, 贪心反例->某行像这样,4 1 1 1 1 1 233 3 3
    //2^80数据, 所以记得高精.
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<cstdlib>
    #include<cstdio>
    using namespace std;
    typedef long long LL;
    typedef __int128 LLL;
    const int maxn = 110;
    int n, m, a[maxn];
    LLL t[maxn], f[maxn][maxn], _max, ans;
    void print(LLL ans){
    if(ans == 0)return ;
    else print(ans/10);
    putchar(ans%10+'0');
    }
    int main(){
    cin>>n>>m;
    t[0] = 1;
    for(int i = 1; i <= m; i++)t[i] = t[i-1]*2;
    while(n--){
    for(int i = 1; i <= m; i++)cin>>a[i];
    memset(f, 0, sizeof f);
    //f[i][j]:这行还剩下[i,j]时能得到的最高分
    //转移加上分别取了左边的和右边的数的时候的得分
    //转移顺序,区间从大到小。
    for(int i = 1; i <= m; i++)
    for(int j = m; j >= i; j--)
    f[i][j]=max(f[i-1][j]+t[m-(j-i+1)]*a[i-1], f[i][j+1]+t[m-(j-i+1)]*a[j+1]);
    //枚举最后一个取的是哪个数,得到这一行的最高分
    _max = 0;
    for(int i = 1; i <= m; i++)_max = max(_max, f[i][i]+t[m]*a[i]);
    ans += _max;
    }
    if(ans == 0)cout<<"0\n";
    else print(ans);
    return 0;
    }

0x06棋盘型动态规划

  1. 过河卒

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    //填表法
    #include<iostream>
    using namespace std;
    int n, m, x, y, a[20][20], f[20][20];
    int main(){
    cin>>n>>m>>x>>y; x++;y++;n++;m++;//+1方便赋初始值
    //9 points could'n be find
    a[x][y] = 1;
    a[x-1][y-2] = a[x-1][y+2] = a[x+1][y-2] = a[x+1][y+2] = 1;
    a[x-2][y-1] = a[x-2][y+1] = a[x+2][y-1] = a[x+2][y+1] = 1;
    f[0][1] = 1; //边界
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= m; j++)
    if(!a[i][j])f[i][j] = f[i-1][j]+f[i][j-1];
    cout<<f[n][m]<<"\n";
    return 0;
    }
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    //刷表法
    #include<iostream>
    using namespace std;
    int n, m, x, y, a[20][20], f[20][20];
    int main(){
    cin>>n>>m>>x>>y;
    a[x][y] = 1;
    a[x-1][y-2] = a[x-1][y+2] = a[x+1][y-2] = a[x+1][y+2] = 1;
    a[x-2][y-1] = a[x-2][y+1] = a[x+2][y-1] = a[x+2][y+1] = 1;
    f[0][0] = 1;//初始值, 刷表时不会覆盖所以可以直接放
    for(int i = 0; i <= n; i++){
    for(int j = 0; j <= m; j++){
    if(!a[i+1][j])f[i+1][j] += f[i][j];
    if(!a[i][j+1])f[i][j+1] += f[i][j];
    }
    }
    cout<<f[n][m]<<"\n";
    return 0;
    }
  2. 传纸条

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    //考虑题设,找到两条不重复的路径,所以从上到下直接DP,状态四维(上往下,下往上分别DP,没办法考虑路径重叠)
    //f[i][j][k][l]表示分别到(i,j),(k,l)时候的最大好心值
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, m, a[55][55], f[55][55][55][55];
    int main(){
    cin>>n>>m;
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= m; j++)
    cin>>a[i][j];
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= m; j++)
    for(int k = 1; k <= n; k++)
    for(int l = 1; l <= m; l++)
    if(!(i==k&&j==l) || (i==n&&j==m&&k==n&&l==m))
    f[i][j][k][l] = max(max(f[i-1][j][k-1][l], f[i][j-1][k-1][l]), max(f[i-1][j][k][l-1],f[i][j-1][k][l-1]))+a[i][j]+a[k][l];
    cout<<f[n][m][n][m]<<"\n";
    return 0;
    }
  3. 骑士游历

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    //bugs:行列弄反(x,y是坐标轴...)+longlong
    #include<iostream>
    using namespace std;
    typedef long long LL;
    LL n, m, x1, y1, x2, y2, f[55][55];
    int main(){
    cin>>n>>m>>x1>>y1>>x2>>y2;
    f[x1][y1] = 1;
    for(int i = x1+1; i <= x2; i++)//避免覆盖掉x1,y1时候的1种方案
    for(int j = 1; j <= n; j++)//因为日字,所以要全
    f[i][j] = f[i-1][j-2]+f[i-1][j+2]+f[i-2][j-1]+f[i-2][j+1];
    cout<<f[x2][y2]<<"\n";
    return 0;
    }
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    //行列式反的,,,
    #include<iostream>
    using namespace std;
    typedef long long LL;
    LL n, m, x1, y1, x2, y2, f[55][55];
    int main(){
    cin>>n>>m>>x1>>y1>>x2>>y2;
    f[y1][x1] = 1;
    //转移的时候按照列每层去取(因为只能往右走)
    for(int i = x1; i <= x2; i++){//枚举y
    for(int j = 1; j <= m; j++){//枚举x
    //之前的状态无法到达那么当前状态也无法到达
    if(!f[j][i])continue;
    //刷表状态转移
    f[j+1][i+2] += f[j][i];
    f[j-1][i+2] += f[j][i];
    f[j+2][i+1] += f[j][i];
    f[j-2][i+1] += f[j][i];
    }
    }
    cout<<f[y2][x2]<<"\n";
    return 0;
    }
  4. 数字三角形

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    //f[i][j]:从(i,j)出发能获得的最大值 _裸DFS
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a[110][110], f[110][110];
    int dfs(int i, int j){
    if(f[i][j])return f[i][j];
    if(i>n || j>n)return 0;
    return f[i][j] = max(dfs(i+1,j),dfs(i+1,j+1))+a[i][j];
    }
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= i; j++)
    cin>>a[i][j];
    cout<<dfs(1,1)<<"\n";
    return 0;
    }
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    //f[i][j]:从(i,j)出发能获得的最大值 _裸递推
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a[110][110], f[110][110];
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= i; j++)
    cin>>a[i][j];
    for(int i = n; i >= 1; i--)
    for(int j = 1; j <= i; j++)
    f[i][j] = max(f[i+1][j], f[i+1][j+1])+a[i][j];
    cout<<f[1][1]<<"\n";
    return 0;
    }
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    //f[i][j]:从(1,1)到(i,j)能获得的最大值 _裸递推
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a[110][110], f[110][110];
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= i; j++)
    cin>>a[i][j];
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= i; j++)
    f[i][j] = max(f[i-1][j], f[i-1][j-1])+a[i][j];
    int ans = -0xffffff;
    for(int i = 1; i <= n; i++)ans = max(ans, f[n][i]);
    cout<<ans<<'\n';
    return 0;
    }
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    //f[i][j]:从(i,j)出发能获得的最大值 _滚动数组
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a[110][110], f[2][110];
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= i; j++)
    cin>>a[i][j];
    for(int i = n; i >= 1; i--)
    for(int j = 1; j <= i; j++)
    f[i%2][j] = max(f[(i+1)%2][j], f[(i+1)%2][j+1])+a[i][j];
    cout<<f[1][1]<<"\n";
    return 0;
    }
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    //f[i][j]:从(1,1)到(i,j)能获得的最大值 _滚动数组2
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, a, f[2][110];
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++){
    for(int j = 1; j <= i; j++){
    cin>>a;
    f[i%2][j] = max(f[(i-1)%2][j], f[(i-1)%2][j-1])+a;
    }
    }
    int ans = -0xffffff;
    for(int i = 1; i <= n; i++)ans = max(ans, f[n%2][i]);
    cout<<ans<<"\n";
    return 0;
    }

0x07划分型动态规划

  1. 乘积最大

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    //f[i][j]:前i位数包含j个乘号时能获得的最大值
    //转移,枚举每个乘号的位置即可,O(n^3)可过。
    #include<iostream>
    #include<algorithm>
    #include<string>
    using namespace std;
    typedef long long LL;
    int n, m;
    LL f[110][110];
    string s;
    LL mid(int l, int r){
    LL t = 0;
    for(int i = l; i <= r; i++)
    t = t*10+s[i-1]-'0';//第i位在s[i-1];
    return t;
    }
    int main(){
    cin>>n>>m>>s;
    for(int i = 1; i <= n; i++)f[i][0] = mid(1,i);//边界条件,没有乘号
    for(int i = 1; i <= n; i++) //枚举前i位数
    for(int j = 1; j <= min(m,i-1); j++)//枚举每个乘号(即子状态)
    for(int k = j; k < i; k++)//枚举该乘号的位置,乘号放后面(保证第j个乘号时, 前j-1个乘号的最优状态已经算出来了)
    f[i][j] = max(f[i][j], f[k][j-1]*mid(k+1,i));
    cout<<f[n][m]<<"\n";
    return 0;
    }
  2. 数的划分

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    //step1:把n个苹果放到m个盘子里,不允许有空盘。等价于每个盘子放一个苹果先允许有空盘
    //step2:f[i][j]表示i个苹果j个盘子的放法数目
    //step3:转移,j>i时,去掉空盘不影响结果; j<=i时,对盘子是否空着分类讨论;
    #include<iostream>
    using namespace std;
    int n, m, f[210][10];
    int main(){
    cin>>n>>m;
    n-=m;//每个盘子先放一个苹果就不会有空盘了。。。
    for(int i = 0; i <= n; i++)f[0][i]=f[i][1]=1;
    for(int i = 1; i <= n; i++)
    for(int j = 2; j <= m; j++)
    f[i][j] = j>i?f[i][i]:f[i][j-1]+f[i-j][j];//所有盘子有苹果时每个盘子都去掉一个苹果不影响结果
    cout<<f[n][m]<<"\n";
    return 0;
    }
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    //Step1:f[i][j]:将i这个整数划分成j份且不重复的方法数
    //Step2:因为划分成的每一份至少为1,所以我们把它每份减去1
    //Step3:将i这个数划分成j份等价于将i-j这个数划分成1份、2份、3份。。。j份的和
    //Step4:f[i-1][j-1]=f[(i-1)-(j-1)][1]+...+f[(i-1)-(j-1)][j-1]; 代入化简
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n, k, f[210][10];
    int main(){
    cin>>n>>k;
    f[0][0] = 1;
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= min(i,k); j++)
    f[i][j] = f[i-j][j]+f[i-1][j-1];
    cout<<f[n][k];
    return 0;
    }
  3. 统计单词个数

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    //just for test4
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<string>
    using namespace std;
    int n, m, x, d[210], f[210][50];//c[210][210];
    string s, w[20];
    void pre(){
    memset(d, 0x3f, sizeof d);
    for(int i = 1; i <= s.size(); i++){//枚举每个字母
    for(int j = 1; j <= x; j++){//枚举每个单词
    if(s.substr(i).find(w[j])==0){//如果存在i字母开头的单词
    d[i] = min(d[i], i+(int)w[j].size()-1);
    }
    }
    }
    }
    int main(){
    int T; cin>>T;
    while(T--){
    //input
    cin>>n>>m;
    s = " ";
    for(int i = 1; i <= n; i++){
    string t; cin>>t; s += t;
    }
    n *= 20;
    cin>>x;
    for(int i = 1; i <= x; i++)cin>>w[i];
    //预处理1:得到c[i][j]为[i,j]中的最大单词数
    //预处理2:得到d[i]为字母i开头的最短单词的结束位置(小贪心,每个字母只能按照第一个字母取一次)
    pre();
    //dp:f[i][j]为前i个字母划分成j段能得到的最大单词数
    //转移:f[i][j] = max{ f[k][j-1]+c[k+1][i] | k>=j&&k<i}
    for(int i = 1; i <= n; i++){
    for(int j = 1; j <= m; j++){
    int w = 0;
    for(int k = i; k >= j; k--){ //逆序覆盖
    if(d[k]<=i)w++; //w=[k,i]
    f[i][j] = max(f[i][j],f[k-1][j-1]+w);
    //f[i][j] = max(f[i][j], f[k][j-1]+w);
    //if(d[k]<=i)w++; //w=[k+1,i]
    }
    }
    }
    cout<<f[n][m]<<"\n";
    }
    return 0;
    }

0x08宽度优先搜索

  1. 四子连棋

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    //思路:把空白当棋,交替黑白走。
    //实现:BFS, 打表判断是否成立
    #include<iostream>
    #include<algorithm>
    #include<string>
    #include<queue>
    using namespace std;
    string s;
    struct node{
    string ma; int step; char next;
    node(string x, int y, char ch):ma(x),step(y),next(ch){}
    };
    queue<node>q;
    int dz[] = {4,-4,1,-1};
    char change(char ch){
    if(ch == 'B')return 'W';
    if(ch == 'W')return 'B';
    }
    int check(string s){
    //check diagonal 1
    if(s[0]==s[5] && s[5]==s[10] && s[10]==s[15])return 1;
    //check diagonal 2
    if(s[3]==s[6] && s[6]==s[9] && s[9]==s[12])return 1;
    //check row
    for(int i = 0; i < 4; i++){
    int ok = 1, t = 4*i;
    for(int j = 0; j < 4; j++)
    if(s[t] != s[t+j])ok = 0;
    if(ok)return 1;
    }
    //check col
    for(int i = 0; i < 4; i++){
    int ok = 1, t = i;
    for(int j = 0; j < 4; j++)
    if(s[t] != s[t+j*4])ok = 0;
    if(ok)return 1;
    }
    return 0;
    }
    int bfs(){
    while(q.size()){
    string t = q.front().ma;
    int st = q.front().step;
    char ch = q.front().next;
    q.pop();
    //check
    if(check(t))return st;
    //find O
    int o1=-1, o2;
    for(int i = 0; i < 16; i++){
    if(t[i]=='O'){
    if(o1==-1)o1 = i;
    else o2 = i;
    }
    }
    //o1go
    for(int i = 0; i < 4; i++){
    if(dz[i]==1 && o1%4==3)continue;
    if(dz[i]==-1 && o1%4==0)continue;
    int nz = o1+dz[i];
    if(nz>=0 && nz<16 && t[nz]==ch){
    string nt = t;
    swap(nt[o1],nt[nz]);
    q.push(node(nt,st+1,change(ch)));
    }
    }
    //o2go
    for(int i = 0; i < 4; i++){
    if(dz[i]==1 && o2%4==3)continue;
    if(dz[i]==-1 && o2%4==0)continue;
    int nz = o2+dz[i];
    if(nz>=0 && nz<16 && t[nz]==ch){
    string nt = t;
    swap(nt[o2],nt[nz]);
    q.push(node(nt,st+1,change(ch)));
    }
    }
    }
    }
    int main(){
    ios::sync_with_stdio(false);
    for(int i = 0; i < 4; i++){
    string t; cin>>t; s += t;
    }
    if(check(s)){ cout<<"0"; return 0;}
    int ans = 0xffffff;
    q.push(node(s,0,'W'));
    ans = min(ans, bfs());
    while(q.size())q.pop();
    q.push(node(s,0,'B'));
    ans = min(ans, bfs());
    cout<<ans<<"\n";
    return 0;
    }
  2. 逃跑的拉尔夫

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    #include<cstdio>
    #include<queue>
    #include<set>//set判重防MLE
    using namespace std;
    const int dx[4] = {-1, 0, 1, 0};
    const int dy[4] = {0, 1, 0, -1};
    int r, c, c1, r1, n, go[1010];
    char a[55][55];
    struct node{
    int x, y, step;
    node(int x, int y, int step):x(x),y(y),step(step){}
    };
    queue<node>q;
    set<int>s;
    bool inside(int x, int y){ return (x<r && x>=0 && y<c && y>=0 && a[x][y]!='X'); }
    int togo(char ch){
    if(ch == 'N')return 0;
    if(ch == 'E')return 1;
    if(ch == 'S')return 2;
    if(ch == 'W')return 3;
    }
    void bfs(){
    q.push(node(r1, c1, 0));
    while(q.size()){
    node t = q.front(); q.pop();
    if(t.step == n)a[t.x][t.y] = '*';
    int tt=go[t.step], nx=t.x+dx[tt], ny=t.y+dy[tt];
    while(inside(nx,ny)){
    int ok = nx*1000000+ny*10000+t.step+1;
    if(!s.count(ok)){
    q.push(node(nx,ny,t.step+1));
    s.insert(ok);
    }
    nx += dx[tt], ny += dy[tt];
    }
    }
    return ;
    }
    int main(){
    scanf("%d%d", &r, &c);
    for(int i = 0; i < r; i++)scanf("%s", a[i]);
    for(int i = 0; i < r; i++)
    for(int j = 0; j < c; j++)
    if(a[i][j] == '*'){ r1 = i; c1 = j; break;}
    a[r1][c1] = '.';
    scanf("%d",&n);
    for(int i= 0; i < n; i++){
    char t[10]; scanf("%s", t); go[i] = togo(t[0]);
    }
    bfs();
    for(int i = 0; i < r; i++)printf("%s\n", a[i]);
    return 0;
    }
  3. 字串变换

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    //思路就是对于每个状态下的字符串,枚举可以替换的部分替换作为下一个新的状态。
    #include<iostream>
    #include<queue>
    #include<string>
    #include<map>
    using namespace std;
    int n = 1, flag;
    string a, b, ai[1010], bi[1010];
    queue<string>q;
    map<string, int>ma;//map判重防MLE
    int main(){
    cin>>a>>b;
    while(cin>>ai[n]>>bi[n])n++;
    q.push(a);
    ma[a] = 0;
    while(q.size()){
    string t = q.front(); q.pop();
    if(t == b){ flag = 1; break; }
    if(ma[t]>10)break;
    //如果没有这层循环的话,就只能找到第一个子串,后面的会被忽略,如abaaaba abcdaba
    for(int j = 0; j < t.size(); j++){
    string nt = t.substr(j);
    for(int i = 1; i < n; i++){
    int tt = nt.find(ai[i]);
    if(tt == string::npos)continue;
    tt += j; //边界条件调起来很麻烦,以及最后直接+j就好了
    string ttt = t.substr(0,tt)+bi[i]+t.substr(tt+ai[i].size());
    if(!ma.count(ttt)){
    ma[ttt] = ma[t]+1;
    q.push(ttt);
    }
    }
    }
    }
    if(flag)cout<<ma[b]<<"\n";
    else cout<<"NO ANSWER!\n";
    return 0;
    }
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    #include<iostream>
    #include<string>
    #include<queue>
    #include<set> //set判重
    #define maxn 1010
    using namespace std;
    string ai[maxn],bi[maxn];
    struct node{
    string str;
    int st;
    node(string a, int b):str(a),st(b){}
    };
    set<string>s;
    int ans;
    int main(){
    string a, b;
    cin>>a>>b;
    int n = 1;
    while(cin>>ai[n]>>bi[n])n++;
    queue<node>q;
    q.push(node(a,0));
    while(q.size()){
    node t = q.front(); q.pop();
    if(t.str == b){ ans = t.st; break;}
    if(t.st > 10){ ans = 20; break; }
    for(int j = 0; j < t.str.size(); j++){
    string nt = t.str.substr(j);
    for(int i = 1; i < n; i++){
    int tt = nt.find(ai[i]);
    if(tt==string::npos)continue;
    tt += j;
    string ttt = t.str.substr(0,tt)+bi[i]+t.str.substr(tt+ai[i].size());
    if(!s.count(ttt)){
    q.push(node(ttt,t.st+1));
    s.insert(ttt);
    }
    }
    }
    }
    //如果<10就因为找不到出来也是不成立的, 即ans没有被赋过值的话
    if(ans == 20 || ans == 0)cout<<"NO ANSWER!\n";
    else cout<<ans<<"\n";
    return 0;
    }

0x09深度优先搜索

  1. 单词接龙

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    //直接搜索就好啦,几乎没什么技巧,就是状态建模会有点难想到(应该有多种)
    //包含的情况可以证明是不需要考虑的,因为包含后一定不会比不包含要来的长
    #include<iostream>
    #include<algorithm>
    #include<string>
    using namespace std;
    const int maxn = 30;
    int n, ans, used[maxn];
    string w[maxn];
    //找到a的末尾与b的前端重复的子串并返回其长度
    int find(string a, string b){
    int mm = min(a.size(), b.size());
    for(int i = 1; i <= mm; i++)
    if(a.substr(a.size()-i)==b.substr(0,i))
    return i;
    return 0;
    }
    //深度优先搜索寻找解, 状态:s为当前字符串
    void dfs(string s){
    ans = max(ans, (int)s.size());
    for(int i = 1; i <= n; i++)if(used[i]<2){
    int t = find(s, w[i]);
    if(t == s.size() && s!=w[0])continue;//包含关系
    if(t){
    used[i]++;
    dfs(s.substr(0,s.size()-t)+w[i]);
    used[i]--;
    }
    }
    }
    int main(){
    cin>>n;
    for(int i = 1; i <= n; i++)cin>>w[i];
    cin>>w[0];
    dfs(w[0]);
    cout<<ans<<"\n";
    return 0;
    }
  2. 四色问题

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    //尝试填每个点每种颜色填过去就好啦
    #include<iostream>
    using namespace std;
    int n, e[10][10];
    int c[10], ans;
    void dfs(int cur){
    if(cur == n)ans++;
    else for(int i = 0; i < 4; i++){
    c[cur] = i;
    bool ok = true;
    for(int j = 0; j < cur; j++)//判断和前面的点有没有冲突
    if(e[j][cur] && c[j]==c[cur])//如果联通且同色那就翻车
    { ok = false; break; }
    if(ok){
    dfs(cur+1);
    }
    }
    }
    int main(){
    cin>>n;
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
    cin>>e[i][j];
    dfs(0);
    cout<<ans<<"\n";
    return 0;
    }
  3. 全排列

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    #include<iostream>
    using namespace std;
    int n, c[20];
    void dfs(int cur){
    if(cur == n){
    for(int i = 0; i < n; i++)cout<<c[i]<<" ";
    cout<<"\n";
    }else for(int i = 1; i <= n; i++){
    int ok = 1;
    for(int j = 0; j < cur; j++)
    if(c[j]==i)ok = 0;
    if(ok){
    c[cur] = i;
    dfs(cur+1);
    }
    }
    }
    int main(){
    cin>>n;
    dfs(0);
    return 0;
    }
  4. N皇后问题

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    //c[i]:第i行的皇后放在第几列
    #include<iostream>
    using namespace std;
    int n, c[20], ans;
    void dfs(int cur){
    if(cur > n)ans++;
    else for(int i = 1; i <= n; i++){
    int ok = 1;
    for(int j = 1; j < cur; j++)
    if(c[j]==i || c[j]-j==i-cur || c[j]+j==i+cur)
    { ok = 0; break; }
    if(ok){
    c[cur] = i;
    dfs(cur+1);
    }
    }
    }
    int main(){
    cin>>n;
    dfs(1);
    cout<<ans<<"\n";
    return 0;
    }