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poj 1050

2024-08-02 07:09:35   221人阅读

To the Max
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 41322 Accepted: 21917

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
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>#include<cstdlib>#include<queue>#include<vector>#include<set>using namespace std;int n,a[102][102],v[102][102],maxx,ans;int main(){      while(scanf("%d",&n)!=EOF)      {            maxx=0,ans=0;            for(int i=1;i<=n;i++)                  for(int j=1;j<=n;j++)                  {                       scanf("%d",&a[i][j]);                       v[i][j]=v[i-1][j]+a[i][j];                  }            for(int i=1;i<=n;i++)            {                  for(int j=i;j<=n;j++)                  {                       ans=0;                       for(int k=1;k<=n;k++)                       {                          if(ans>0)                            ans+=v[j][k]-v[i-1][k];                          else                            ans=v[j][k]-v[i-1][k];                          if(ans>maxx)                              maxx=ans;                       }                  }            }            printf("%d\n",maxx);      }      return 0;}

　　

poj 1050


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