poj 1050 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.

http://poj.org/problem?id=1050

 

代码：计算从第i行到第j行每一列的和(i,j属于0,,,,,,n-1)，最后构成一个一维数组，继而求该一维数组的最长子序列和，即为该第i行到第j列的最大矩阵和

#include<stdio.h>#include<stdlib.h>#include<string.h>#define max 105int main(){int n;while(scanf("%d",&n) != EOF){int sum = -65536;int dp[n + 1][n + 1];int arr[n + 1][n + 1];memset(dp,0,sizeof(dp));for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)scanf("%d",&arr[i][j]);for(int k = 1; k <= n; k++){for(int i = k; i <= n; i++){int tmpsum = 0 , maxsum;for(int j = 1; j <= n; j++){dp[i][j] = dp[i - 1][j] + arr[i][j];int tmp = dp[i][j];if(j == 1)maxsum = tmp;tmpsum += tmp;if(tmpsum > maxsum)maxsum = tmpsum;else if(tmpsum < 0)tmpsum = 0;}if(maxsum > sum)sum = maxsum;}memset(dp,0,sizeof(dp));}printf("%d\n",sum);}}