POJ_1050_(dp)

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48232		Accepted: 25534

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

---

给一个矩阵，求其子矩阵所有元素加和的最大值。。。

没想到n^4的复杂度能过。。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 105
#define INF 999999999

int num[N][N];
int dpUL[N][N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            scanf("%d",&num[i][j]);
    int res=-INF;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            dpUL[i][j]=dpUL[i-1][j]+dpUL[i][j-1]-dpUL[i-1][j-1]+num[i][j];
            res=max(res,dpUL[i][j]);
            for(int r=0; r<i; r++)
            {
                for(int c=0; c<j; c++)
                    res=max(res,dpUL[i][j]-dpUL[i][c]-dpUL[r][j]+dpUL[r][c]);
            }
        }
    printf("%d
",res);
    return 0;
}
/*
-1 -1 -1 -1
-1 2 2 -1
-1 2 2 -1
-1 -1 -1 -1
*/