POJ 1050 To the Max 最大子矩阵和（二维的最大字段和）

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:

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

Greater New York 2001

 

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 ;
}