最大子矩阵和

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

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

8  0 -2

Sample Output

15

DP 求最大和子矩阵 , 可以用最大和连续子序列的思路解 .

首先 , 读数据的时候 a[i][j] 为前 i 行数据的第 j 个数的和 , 当求第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵的时候, 就以 p 行和 q 行之间的同一列的数字的和作为一个数列的元素 , 然后 DP 求这个数列的最大和连续子序列 , 就是第p 行到第 q 行之间宽为 p-q+1 的最大和矩阵 .

#include<cstdio>
#include<cstring>
using namespace std;
const int INF=999999999;
int n,m;
int d[110][110];
int s[110];
int main(){
    int i,j,k,t;
    while(scanf("%d",&n)!=EOF){
        for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&d[i][j]);
        m=-INF;
        for(i=1;i<=n;i++){
            memset(s,0,sizeof(s));
            for(j=i;j<=n;j++){
                t=0;
                for(k=1;k<=n;k++){
                    s[k]+=d[j][k];
                    if(t<=0) t=s[k];
                    else t+=s[k];
                    if(t>m)m=t;
                }
            }
        }
        printf("%d\n",m);
    }
return 0;
}