Ural 1146 Maximum Sum

Problem Description

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.

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-hand corner and has the 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 white-space (newlines and spaces). These N ² integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

 

Sample Input

input	output
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2	15

 

 输入矩阵mat[i][j]：

0  -2  -7  0

9   2  -6  2

-4  1  -4  1

-1  8  0  -2

sum[i][j]表示从mat[1][j]到mat[i][j]的总和：

0  2  -7   0

9  4 -13  2

5  5 -17  3

4 13 -17 1

这样可以枚举出从下到上的每一列的和。

for(int r1=1; r1<=n; r1++){
　　for(int r2=r1; r2<=n; r2++){

　　　　//循环每一列这样就可以遍历每一个子阵了。

　　　　//求的方法跟求子段的和一样

　　}

}

#include <stdio.h>
#include <iostream>
#define MAXN 110
using namespace std; 

int n;
int mat[MAXN][MAXN];
int sum[MAXN][MAXN];

int main()
{
    while( scanf("%d",&n)!=EOF ){
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                scanf("%d",&mat[i][j]);
            }
        }
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                if(i==1){
                    sum[i][j]=mat[i][j];
                }else{
                    sum[i][j]=sum[i-1][j]+mat[i][j];    
                }        
            }
        }
        int ans=-999;
        for(int r1=1; r1<=n; r1++){
            for(int r2=r1; r2<=n; r2++){
                int temp=0;
                for(int j=1; j<=n; j++){
                    temp+=sum[r2][j]-sum[r1-1][j];
                    ans=max( temp ,ans );
                    if(temp<0)temp=0;
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}