传送门：
http://poj.org/problem?id=1050
To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 52306 | Accepted: 27646 |
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:
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
Source
分析：
给你一个N*N的数字矩阵
问你子矩阵的最大和是多少
非常的类似最大子段和问题
dp[i][j]:表示从第i列到j列的子矩阵的最大和
那么在第i列到第j列中
第一行的和就看成一个一个数
第二行的和也是看成一个数
第n行的和也是看成一个数
在这些一维线性的数里面找最大的子段和
比如样例：
第列到第四列
（-2-7+0）=-9
（2-6+2）=-4
（1-4+1）=-2
（8+0-2）=6
在-9，-4，-2，6里面找最大的子段和，看出来是6
那么6就是2列到4列的最大子矩阵和
i列：从1到n
j列：从i到n
code：
#include <iostream>
#include <cstdio>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<memory>
#include<queue>
#include<vector>
using namespace std;
#define max_v 105
#define INF 99999999
int dp1[max_v][max_v];//起始i列 终止j列的max
int dp2[max_v];//最大子段和的dp，代表以第i个数结尾的最大子合和值
int a[max_v][max_v];
int f(int j1,int j2,int i)//j1列到j2列，i行上数字的和
{
int sum=;
for(int j=j1;j<=j2;j++)
{
sum+=a[i][j];
}
return sum;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
scanf("%d",&a[i][j]);
memset(dp1,,sizeof(dp1));
dp1[][]=a[][];
int result=-INF;
for(int i=;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
memset(dp2,,sizeof(dp2));
dp2[]=f(i,j,);
int maxv=dp2[];
for(int k=;k<=n;k++)
{
int x=;
if(dp2[k-]>)
x=dp2[k-];
dp2[k]=x+f(i,j,k);
maxv=max(maxv,dp2[k]);
}
dp1[i][j]=maxv;
result=max(result,dp1[i][j]);
}
}
printf("%d\n",result);
}
return ;
}