poj To the Max 1050 （二维最大子矩阵 DP）

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 44258 Accepted: 23461

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

#include<stdio.h>#include<string.h>#define MAX(a,b) (a>b?a:b)int map[110][110];int n,m,max;void find(int x){int t=0,i,j;for(i=1;i<=n;i++){if(t>0)t+=map[x][i];elset=map[x][i];max=MAX(max,t);}}int main(){int t,i,j,k;while(scanf("%d",&n)!=EOF){memset(map,0,sizeof(map));for(i=1;i<=n;i++){for(j=1;j<=n;j++){scanf("%d",&map[i][j]);}}max=map[1][1];for(i=1;i<=n;i++){find(i);for(j=i+1;j<=n;j++){for(k=1;k<=n;k++){map[i][k]+=map[j][k];}find(i);}}printf("%d\n",max);}return 0;}