POJ1191 棋盘分割

由于题目是中文的，这里就不解释题目大意。是一个很好的dp题目，不过这个题目需要将平方差公式变形，两次运用dp的思想，对中间结果dp。然后的求出最终结果。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>

using namespace std;

const int MAX = 10;
const int MAXN = 20;
const int INF = 1000000000;

int a[MAX][MAX],d[MAXN][MAX][MAX][MAX][MAX]; //d树组表示还剩余k次切割时对应的棋盘（这里棋盘用相应的左上方和右下方的顶点表示）

int sum(int x1, int y1, int x2, int y2){           //a[i][j]表示棋盘(1,1)到(i,j)的所有权值的和。
    return a[x2][y2] - a[x1-1][y2] - a[x2][y1-1] + a[x1-1][y1-1];
}

int dp(int k, int x1, int y1, int x2, int y2){
    if (d[k][x1][y1][x2][y2] != -1){
        return d[k][x1][y1][x2][y2];
    }

    int s1, s2;
    if (k == 0){
        s1 = sum(x1, y1, x2, y2);
        return d[k][x1][y1][x2][y2] = s1*s1;
    }

    int minn = INF, tmp1;
    for (int i = x1; i<x2; i++){
        s1 = sum(i+1, y1, x2, y2);
        s2 = sum(x1, y1, i, y2);
        tmp1 = min(dp(k-1, x1, y1, i, y2) + s1*s1, dp(k-1, i+1,y1, x2, y2) + s2*s2);
        minn = min(minn, tmp1);
    }

    int tmp2 = INF;
    for (int i = y1; i<y2; i++){
        s1 = sum(x1, i+1, x2, y2);
        s2 = sum(x1, y1, x2, i);
        tmp2 = min(dp(k-1, x1, y1, x2, i) + s1*s1, dp(k-1,x1, i+1, x2, y2) + s2*s2);
        minn = min(tmp2, minn);
    }
    
    return d[k][x1][y1][x2][y2] = minn;
}

int main(){
    int n;
    while (scanf("%d",&n) != EOF){
        memset(d, -1, sizeof(d));
        memset(a, 0,sizeof(a));
        int tmp = 0, t;
        for (int i = 1; i<=8; i++)
            for (int j = 1; j<=8; j++){
                scanf("%d",&t);
                tmp += t;
                a[i][j] = t + a[i-1][j] + a[i][j-1] - a[i-1][j-1];
            }

        int ans = dp(n-1, 1,1,8,8);

        double res = tmp*1.0 / n; 


        printf("%.3lf\n",sqrt(ans*1.0/n -res*res));
    }

    return 0;
}