杭电ACM HDU 1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6909    Accepted Submission(s): 3328

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

   
   
   
   

    
    
    
    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

 

#include <stdio.h>
int main(){
    int n,x;
    int arr[100*101];
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d",&x);
                //表示第i行前j个数之和
                arr[i*n+j]=x+(j?arr[i*n+j-1]:0);
            }
        }
        int max=-128;
        for(int i=0;i<n;i++){
            for(int j=i;j<n;j++){
                for(int k=0,t=0;k<n;k++){
                    //tt表示第k行第j列与第i列之间的数
                    int tt=t+arr[k*n+j]-(i?arr[k*n+i-1]:0);
                    t=tt>0?tt:0;
                    if(max<t)max=t;
                }
            }
        }
        printf("%d\n",max);
    }
}