UASCO Count the rectangles

Count the rectangles

TimeLimit: 2 Second   MemoryLimit: 64 Megabyte

Totalsubmit: 64   Accepted: 36  

Description

You are given numbers of rectangles made of '1's and '0's.Calculate how many small rectangles which have four '1's on the four corners.

Input

There are several test cases, each test case starts with a line containing two positive integers n and m. n and m is the size of the rectangle (1<=n<=100, 1<=m<=100). Next follow a rectangle which contains only number '0' and '1'. The input will finish with the end of file.

Output

The number of rectangles which meet the requirements.

Sample Input

3 4
1 0 1 0
0 1 1 0
1 1 1 0

Sample Output

2

Source

zzx

 

思路：

 

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
int map[110][110];
int the_last_sum;
int the_long_x;
int the_long_y;
int n,m;
int main()
{
    while(scanf("%d%d",&n,&m) != EOF)
    {
        memset(map,0,sizeof(map));
        for(int i = 1;i <= n;i ++)
            for(int j = 1;j <= m;j ++)
                scanf("%d",&map[i][j]);
        the_last_sum = 0;
        for(int i = 1;i <= n;i ++)
            for(int j = 1;j <= m;j ++)
            {
                 the_long_x = 0;
                 the_long_y = 0;
                 if(map[i][j] == 1 && j + 1 <= m && i + 1 <= n)
                 {
                     for(int k = j + 1;k <= m;k ++)
                     {
                         if(map[i][k] == 1)
                         {

                             the_long_y = k - j;
                             for(int l = i + 1;l <= n;l ++)
                             {
                                if(map[l][j] == 1)
                                {
                                      the_long_x = l - i;
                                      if(the_long_x != 0 && the_long_y != 0)
                                      {
                                           if(i + the_long_x <= n && j + the_long_y <= m
                                           && map[i + the_long_x][j + the_long_y] == 1)
                                                       the_last_sum ++;
                                      }
                                }
                             }
                         }
                     }
                 }
            }
      printf("%d\n",the_last_sum);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/GODLIKEING/p/3426710.html