poj 1050 dp

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

求最大子矩阵

自己太弱了，曾经写过一次还wa了n次

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define max(a,b) a>b?a:b
using namespace std;
int a[110][110],n;
int max1(int a[]) {
    int sum=-10000000,b=0;
    for(int i=0; i<n; i++) {
        if(b<0)b=a[i];
        else b+=a[i];
        sum=max(b,sum);
    }
    return sum;
}
int max2() {
    int sum=-10000000;
    for(int i=0; i<n; i++) {
        int b[110]= {0};
        for(int j=i; j<n; j++) {
            for(int k=0; k<n; k++) {
                b[k]+=a[j][k];
            }
            sum=max(max1(b),sum);//这里错了好久
        }
    }
    return sum;
}
int main() {
    while(cin>>n) {
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                scanf("%d",&a[i][j]);
        printf("%d\n",max2());
    }
    return 0;
}