【poj 1050】 To the Max 【Greater New York 2001】

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

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

8  0 -2

Sample Output

15

这道题是一个DP，求最大子矩阵和(尽管读入一点也不像一个矩阵)，可以使用和数组将二维的矩阵压缩到一维，然后就是简单的最大子段和了，下面是程序：

#include<stdio.h>
#include<iostream>
using namespace std;
const int INF=0x7fffffff;
int map[105][105],sum[105][105];
int read(){
	char c=getchar();
	int s=0,f=1;
	while((c<'0'||c>'9')&&c!='-'){
		c=getchar();
	}
	if(c=='-'){
		f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		s*=10;
		s+=c-'0';
		c=getchar();
	}
	return s*f;
}
int main(){
	int n=read(),i,j,k,ans=-INF;
	for(i=1;i<=n;i++){
		for(j=1;j<=n;j++){
			map[i][j]=read();
			sum[i][j]=sum[i-1][j]+map[i][j];
		}
	}
	for(i=1;i<=n;i++){
		for(j=i;j<=n;j++){
			int s=-INF,tp=0;
			for(k=1;k<=n;k++){
				int x=sum[j][k]-sum[i-1][k];
				tp+=x;
				if(tp<0){
					tp=0;
				}
				s=max(tp,s);
			}
			ans=max(ans,s);
		}
	}
	printf("%d\n",ans);
	return 0;
}