HDU1081

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10498    Accepted Submission(s): 5043

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 -18 0 -2

 

Sample Output

15把每一列的和累加起来，然后遍历任意两行的差值，在按照最大子段和来计算就可以得出想要的结果
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAX=104;int a[MAX][MAX];int dp[MAX];int main(){int n;while(~scanf("%d",&n)){memset(a,0,sizeof a);memset(dp,0,sizeof dp);for(int i=1; i<=n; i++){for(int j=1; j<=n; j++){scanf("%d",&a[i][j]);a[i][j]+=a[i-1][j];}}int max_sum=0;for(int i=1; i<=n; i++){for(int j=0; j<i; j++){for(int k=1; k<=n; k++){dp[k]=a[i][k]-a[j][k];dp[k]+=dp[k-1];if(dp[k]<0)dp[k]=0;if(dp[k]>max_sum)max_sum=dp[k];}}}printf("%d\n",max_sum);}}