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

　　}

}

 

 1 #include <stdio.h>
 2 #include <iostream>
 3 #define MAXN 110
 4 using namespace std; 
 5 
 6 int n;
 7 int mat[MAXN][MAXN];
 8 int sum[MAXN][MAXN];
 9 
10 int main()
11 {
12     while( scanf("%d",&n)!=EOF ){
13         for(int i=1; i<=n; i++){
14             for(int j=1; j<=n; j++){
15                 scanf("%d",&mat[i][j]);
16             }
17         }
18         for(int i=1; i<=n; i++){
19             for(int j=1; j<=n; j++){
20                 if(i==1){
21                     sum[i][j]=mat[i][j];
22                 }else{
23                     sum[i][j]=sum[i-1][j]+mat[i][j];    
24                 }        
25             }
26         }
27         int ans=-999;
28         for(int r1=1; r1<=n; r1++){
29             for(int r2=r1; r2<=n; r2++){
30                 int temp=0;
31                 for(int j=1; j<=n; j++){
32                     temp+=sum[r2][j]-sum[r1-1][j];
33                     ans=max( temp ,ans );
34                     if(temp<0)temp=0;
35                 }
36             }
37         }
38         printf("%d\n",ans);
39     }
40     return 0;
41 }