To The Max

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. 

As an example, the maximal sub-rectangle of the array: 

0 -2 -7 0 
9 2 -6 2 
-4 1 -4 1 
-1 8 0 -2 

is in the lower left corner: 

9 2 
-4 1 
-1 8 

and has a sum of 15. 

Input

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range −127,127−127,127. 

Output

Output the sum of the maximal sub-rectangle. 

Sample Input

40 -2 -7 0 

9  2 -6 2

-4 1 -4 1

-1 8 0 -2

Sample Output

15

题意：找最大的矩阵和

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n;int maxsum(int f[]){    int dp[101],sum=f[1];    memset(dp,0,sizeof(dp));    for(int i=1;i<=n;i++)    {        dp[i]=max(dp[i],dp[i-1]+f[i]);        sum=max(sum,dp[i]);    }    return sum;}int main(){    int a[101][101];    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%d",&a[i][j]);                a[i][j]=a[i-1][j]+a[i][j];//将前面的每一行累加的下一行            }        }        int s[101],maxx=a[1][1];        for(int i=1;i<=n;i++)//遍历每一个区域        {            memset(s,0,sizeof(s));            for(int j=i;j<=n;j++)            {                for(int l=1;l<n;l++)                    s[l]=a[j][l]-a[i-1][l];                maxx=max(maxx,maxsum(s));            }        }        printf("%d\n",maxx);    }    return 0;}