hdu1081 To the Max

To the Max

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) : 1

Problem 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

求最大矩阵和

做dp[i][j]表示前i行前j列最大值

则有dp[i][j]=max(dp[i][j],dp[i][j-1]+a[i][j])

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#define INF 0xffff
using namespace std;

int main()
{
int n;
int a[150][150];
int dp[150][150];///dp[i][j]表示在i行时前j列的最大值
while(~scanf("%d",&n))
{
memset(dp,0,sizeof(dp));
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
scanf("%d",&a[i][j]);
dp[i][j]+=dp[i][j-1]+a[i][j];
}
//dp[0][0]=a[0][0];
int maxn=-100000;
for(int i=1;i<=n;i++)
{
//dp[i][0]=max(dp[i][0],dp[i-1][0]+a[i][j]);
for(int j=i;j<=n;j++)
{
int sum=0;
for(int k=1;k<=n;k++)
{
if(sum<0)
sum=0;
sum+=dp[k][j]-dp[k][i-1];
if(sum>maxn)
maxn=sum;
}
//dp[i][j]=max(dp[i][j],dp[i-1][j]+dp[i][j-1]+a[i-1][j]+a[i][j-1]);
}
}
cout<<maxn<<endl;
}
return 0;
}