poj 1050 To the Max （dp）

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To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47234		Accepted: 25026

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

Source

Greater New York 2001

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题解：dp

f[i][j][k]表示以i,j为右下角，向上延伸k行的最大子矩阵。

f[i][j][k]=max(f[i][j-1][k]+sum[i][j]-sum[i-k][j],sum[i][j]-sum[i-k][j])

其中sum[i][j]表示的是j列前i行的前缀和。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define N 103
using namespace std;
int f[N][N][N],n,m,sum[N][N],a[N][N];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) 
	 for (int j=1;j<=n;j++) scanf("%d",&a[i][j]);
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	  sum[i][j]=sum[i-1][j]+a[i][j];
	memset(f,128,sizeof(f));
	int ans=f[0][0][0];
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	  for (int k=1;k<=i;k++)
	   f[i][j][k]=max(f[i][j-1][k]+sum[i][j]-sum[i-k][j],sum[i][j]-sum[i-k][j]),ans=max(ans,f[i][j][k]);
    printf("%d\n",ans);
}