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): 10594    Accepted Submission(s): 5086

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

 

 

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 const int maxn=200;
 7 int a[maxn][maxn];
 8 
 9 int main()
10 {
11     int n,t,sum,max;
12     while(scanf("%d",&n)!=EOF)
13     {
14         max=0;
15         for(int i=1;i<=n;i++){
16             for(int j=1;j<=n;j++){
17                 scanf("%d",&t);
18                 a[i][j]=a[i-1][j]+t;
19             }
20         }
21 
22         for(int i=1;i<n-1;i++){
23             for(int j=i;j<=n;j++){
24                 sum=0;
25                 for(int k=1;k<=n;k++){
26                     t=a[j][k]-a[i-1][k];
27                     sum+=t;
28                     if(sum<0) sum=0;
29                     if(sum>max) max=sum; 
30                 }
31             }
32         }
33         printf("%d\n",max);
34     }
35 }