2.txt

中南赛区ACM竞赛题 Description Given a two-dimensional array of positive and negative integers, a sub-rec

题目
To the Max

Time Limit:1000MS  Memory Limit:10000K
Total Submit:2053 Accepted:996 

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

Source 
Greater New York 2001


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代码:
#include <iostream.h>

int a[101][101];

int maxsum(int a[],int n)
{
    int sum,b;
    sum=b=0;
    for(int i=1;i<=n;i++)
    {
        if(b>0)
         b+=a[i];
        else
         b=a[i];
        if(b>sum)
         sum=b;
    }    
    return sum;
}

int maxsum2(int m,int n)
{
int sum=0;
int*b;
b=new int[n+1];
for(int i=1;i<=m;i++)
{
     for(int k=1;k<=n;k++)
      b[k]=0;
     for(int j=i;j<=m;j++)
     {
         for(int k=1;k<=n;k++)
           b[k]+=a[j][k];
           int max=maxsum(b,n);
           if(max>sum)
            sum=max;
     }
}     
return sum;
}

int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
      cin>>a[i][j];
      cout<<maxsum2(n,n)<<endl;
//      cin>>n;
    return 0;
}