poj 1050 To the Max 最大子矩阵和 经典dp

To the Max

 

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

思路：经典dp（yan 教的）

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+10,M=1e6+10,inf=1e9+10;
int a[N][N];
int sum[N][N];
int num[N];
int main()
{
    int x,y,z,i,t;
    while(~scanf("%d",&x))
    {
        memset(sum,0,sizeof(sum));
        for(i=1;i<=x;i++)
        for(t=1;t<=x;t++)
        scanf("%d",&a[i][t]);
        for(i=1;i<=x;i++)
        for(t=1;t<=x;t++)
        sum[t][i]=sum[t-1][i]+a[t][i];
        int maxx=-inf;
        for(i=1;i<=x;i++)
        {
            for(t=i;t<=x;t++)
            {
                for(int j=1;j<=x;j++)
                num[j]=sum[t][j]-sum[i-1][j];
                int sum=0;
                for(int j=1;j<=x;j++)
                {
                    sum+=num[j];
                    maxx=max(maxx,sum);
                    if(sum<0)
                    sum=0;
                }
            }
        }
        printf("%d
",maxx);
    }
    return 0;
}