poj1050 To the Max

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

这道题最重要的是把二维数组的问题转化为一维数组的求最大字段和的问题。我们选定i行和j行，上下一压缩，二维就变为一维了，而每一项即为从i行到j行的元素的和。之后求最大字段和即可。
代码如下：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#define MAX 102
#define MIN -12702
int Mat[MAX][MAX];
int sum[MAX][MAX][MAX];
//        col  rowi rowk i < j
/*
    1 2 3 
    4 5 6
    7 8 9

 */
int main(int argc, char const *argv[])
{
    int N;
    scanf("%d",&N);
    for(int i = 0; i < N; i++) {
        for(int j = 0; j < N; j++) {
            scanf("%d",&Mat[i][j]);
            sum[j][i][i] = Mat[i][j];
        }
    }

    for(int j = 0; j < N; j++) {
        for(int i = 0; i < N; i++) {
            for(int k = i+1; k < N; k++) {
                sum[j][i][k] = sum[j][i][k-1] + Mat[k][j];
            }
        }
    }
    int ans = MIN;
    for(int i = 0; i < N; i++) {
        int maxj = MIN;
        for(int j = i; j < N; j++) {
            //from i row to j row
            int ansij = 0;
            int maxij = MIN;
            for(int k = 0; k < N; k++) {
                ansij = ansij + sum[k][i][j];
                if(maxij < ansij) {
                    maxij = ansij;
                }
                if(ansij < 0) {
                    ansij = 0;
                }
            }
            if(maxij > maxj) {
                maxj = maxij;
            }
        }
        if(maxj > ans) {
            ans = maxj;
        }
    }
    printf("%d
",ans);
    //system("pause");
    return 0;
}

这道题开始有些不知道怎么做，也是参考了别人的思路才做出来，不过一遍就AC了