poj 1191

这道题目心得，学会怎么求一个矩阵从（x1,y1)到(x2,y2)的数字的总和，并且还有结合动态规划，

#include"stdafx.h"
#include <stdio.h>  
#include <math.h>  
#include <algorithm>  
  
using namespace std;  
  
int matrix[8][8];  
int s[8][8][8][8];  
int dp[15][8][8][8][8];  
int cnt;  
double sum;  
  
double mmin(double a, double b)  
{  
     return a < b ? a : b;  
}   
  
int main()  
{  
    int x1, x2, y1, y2, k, a;  
    double average, sum = 0;  
    scanf("%d", &cnt);  
    for (x1=0; x1<8; x1++)  
    {  
        for (y1=0; y1<8; y1++)  
        {  
            scanf("%d", &matrix[x1][y1]);  
            sum += matrix[x1][y1];  
        }  
    }  
      
    average = sum/cnt;  
      
    for (x1=0; x1<8; x1++)  
    {  
        for (y1=0; y1<8; y1++)  
        {  
            for (x2=x1; x2<8; x2++)  
            {  
                sum = 0;  
                for (y2=y1; y2<8; y2++)  
                {  
                    sum += matrix[x2][y2];  
                    if (x2 == x1)  
                    {  
                        s[x1][y1][x2][y2] = sum;  
                    }  
                    else  
                    {  
                        s[x1][y1][x2][y2] = s[x1][y1][x2-1][y2] + sum;  
                    }  
                      
                    dp[0][x1][y1][x2][y2] = s[x1][y1][x2][y2] * s[x1][y1][x2][y2];  
                }  
            }  
        }  
    }  
      
    int temp;  
      
    for (k=1; k<=cnt-1; k++)  
    {  
        for (x1=0; x1<8; x1++)  
        {  
            for (y1=0; y1<8; y1++)  
            {  
                for (x2=x1; x2<8; x2++)  
                {  
                    for (y2=y1; y2<8; y2++)  
                    {  
                        dp[k][x1][y1][x2][y2] = 2000000000;  
                        for (a=x1; a<x2; a++)  
                        {  
                            temp = mmin(dp[k-1][x1][y1][a][y2] + s[a+1][y1][x2][y2] * s[a+1][y1][x2][y2],   
                                        dp[k-1][a+1][y1][x2][y2] + s[x1][y1][a][y2] * s[x1][y1][a][y2]);  
                            dp[k][x1][y1][x2][y2] = mmin(dp[k][x1][y1][x2][y2], temp);  
                        }  
                          
                        for (a=y1; a<y2; a++)  
                        {  
                            temp = mmin(dp[k-1][x1][y1][x2][a] + s[x1][a+1][x2][y2] * s[x1][a+1][x2][y2],  
                                        dp[k-1][x1][a+1][x2][y2] + s[x1][y1][x2][a] * s[x1][y1][x2][a]);  
                            dp[k][x1][y1][x2][y2] = mmin(dp[k][x1][y1][x2][y2], temp);  
                        }  
                    }  
                }  
            }  
        }  
    }  
      
    double ret = sqrt((double)dp[cnt-1][0][0][7][7] / (double)cnt - average * average);  
    printf("%.3f\n", ret);

    system("pause");
    return 0;  
}