poj 1050 To the Max （dp）

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal 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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

Source

Greater New York 2001

[Submit]   [Go Back]   [Status]   [Discuss]

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 103using 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);}