hdu 1081 & poj 1050 To The Max(最大和的子矩阵)

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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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

Source

Greater New York 2001

题意：给你一个N*N的矩阵，求其中和最大的子矩阵的值！

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int main(){int i, j, k, t;int a, sum, max, N, m[147][147];while(~scanf("%d",&N)){memset(m,0,sizeof(m));for(i = 1; i <= N; i++){for(j = 1; j <= N; j++){scanf("%d",&a);m[i][j]+=m[i][j-1]+a;//表示第i行前j个数之和 }}max = -128;for(i = 1; i <= N; i++)//起始列 {for(j = i; j <= N; j++)//终止列 {sum = 0;for(k = 1; k <= N; k++)//对每一行进行搜索 {if(sum < 0)sum = 0;sum+=m[k][j]-m[k][i-1];//m[k][j]-m[k][i-1]表示第k行第i列之间的数 if(sum > max)max = sum;}}}printf("%d\n",max);}return 0;}