hdu 1081 To The Max (动态规划)

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5917    Accepted Submission(s): 2807

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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 
-1 8 0 -2

 

Sample Output

15

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define N 110int f[N][N];int main(){int n,i,j,k,tmp;int ans;while(cin>>n){memset(f,0,sizeof(f));for(i=1;i<=n;i++)for(j=1;j<=n;j++){cin>>k;f[i][j]=f[i][j-1]+k;//求和这一列1-j的和}ans=-99999999;for(i=1;i<=n;i++)//枚举1-n列for(j=1;j<=i;j++){tmp=-1;for(k=1;k<=n;k++){//枚举1-n行if(tmp>0)tmp+=f[k][i]-f[k][j-1];//上一个结果大于0就累加进去else tmp=f[k][i]-f[k][j-1];//矩形（k,j-1)-(k,j)的和ans=ans>tmp?ans:tmp;}}cout<<ans<<endl;}return 0;}