POJ 1050 To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 36457 Accepted: 19143

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：

求出所给矩阵的最大子矩阵和

 

代码：

#include<stdio.h>#include<string.h>int a[102][102],b[102][102];int N;void merge(int k)                      //将求最大子矩阵和转化为求最大子段和 分别将相邻行相加{int i,j,l,m;memset(b,0,sizeof(b));for(j=0;j<N;j++){for(i=0;i<N-k+1;i++){m=0;for(l=i;m<k;l++,m++){b[i][j]+=a[l][j];}}}}int main(){int i,j,k,sum,max=-10000000;while(scanf("%d",&N)!=EOF){for(i=0;i<N;i++)for(j=0;j<N;j++)scanf("%d",&a[i][j]);k=1;while(k<=N){merge(k);for(i=0;i<N-k+1;i++){sum=-1;for(j=0;j<N;j++){if(sum<0) sum=b[i][j];else sum+=b[i][j];if(sum>max) max=sum;}}k++;}printf("%d\n",max);}return 0;}

思路：

将求最大子矩阵和转化为求最大字段和 分别将相邻行相加