To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4842 Accepted Submission(s): 2289

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

View Code

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define Max(x,y) x>y?x:y
 4 #define INF -0x7fffffff/2
 5 int max_sub(int x[],int n)
 6 {
 7     int max=x[0];
 8     int sum=0;
 9     int i;
10     for(i=0;i<n;i++)
11     {
12         sum+=x[i];
13         if(sum<0)
14         {
15             sum=0;
16         }
17         if(max<sum)
18         {
19             max=sum;
20         }
21     }
22
23     return max;
24 }
25 int  main()
26 {
27     int t;
28     int i,j,k;
29     int aa[105][105];
30     int bb[105];
31     int maxx;
32
33     while(~scanf("%d",&t))
34     {
35         maxx=INF;
36         for(i=0;i<t;i++)
37         {
38             for(j=0;j<t;j++)
39             {
40                 scanf("%d",&aa[i][j]);
41             }
42         }
43
44         for(i=0;i<t;i++)
45         {
46             memset(bb,0,sizeof(bb));
47             for(j=i;j<t;j++)
48             {
49                 for(k=0;k<t;k++)
50                 {
51                     bb[k]+=aa[j][k];
52                 }
53
54                 maxx=maxx>max_sub(bb,t)?maxx:max_sub(bb,t);
55             }
56         }
57         printf("%d\n",maxx);
58     }
59 }

转载于:https://www.cnblogs.com/1114250779boke/archive/2012/08/05/2624312.html

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