G - To the Max POJ - 1050 最大子矩阵和

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

将二维转换为一维，再用一般的最大连续子串和来更新max；

定义两个指针i和j指向矩阵的列，将第k行从第i列到第j列的元素和表示为sum,现在有k个sum组成一个一维序列，求得该序列最大连续子串和来更新max，这样将所有i和j的值遍历完后就得到max了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[110][110];
int sum[110][110];
int main(){
    int n;
    while(cin>>n){
        memset(a,0,sizeof(a));memset(sum,0,sizeof(sum));
        for(int i=0;i<n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&a[i][j-1]);
                sum[i][j]=sum[i][j-1]+a[i][j-1];
            }
        }
        int ans=INT_MIN;
        for(int i=0;i<n;i++)
            for(int j=i;j<=n;j++)
            {
                int sum1=0;
                for(int k=0;k<n;k++){
                    sum1+=sum[k][j]-sum[k][i];
                    if(sum1<0)sum1=0;
                    else if(sum1>ans)ans=sum1;
                }
            }
        printf("%d\n",ans);
    }
}