hdu1081（最大子矩阵和）

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11151 Accepted Submission(s): 5366

Problem Description
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].

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

题意：求n*n矩阵的最大子矩阵

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;long long a[510][510], c[550];int main(){    int n, m;    while(scanf("%d", &m) != EOF){    for(int i = 0; i < m; i++)    {        for(int j = 0; j < m; j++)        {            scanf("%I64d", &a[i][j]);        }    }    long long x, ans, maxx = -1100000000;    for(int i = 0; i < m; i++)    {        for(int k = 0; k < m; k++)            c[k] = 0;        for(int j = i; j < m; j++)        {            for(int k = 0; k < m; k++)            {                c[k] = c[k] + a[j][k];                x = 0, ans = -1100000000;                for(int r = 0; r <= k; r++)                {                    if(x > 0)                    {                        x += c[r];                    }                    else                    {                        x = c[r];                    }                    if(x > ans)                    {                        ans = x;                    }                }              //  printf("%I64d\n", ans);                maxx = max(ans, maxx);            }        }    }    printf("%I64d\n", maxx);}