hdu4331 Image Recognition

Image Recognition

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 964    Accepted Submission(s): 371

Problem Description

Now there is an image recognition problem for you. Now you are given an image which is a N * N matrix and there are only 0s and 1s in the matrix. And we are interested in the squares in whose four edges there is no 0s. So it’s your task to find how many such squares in the image.

 

Input

The first line of the input contains an integer T (1<=T<=10) which means the number of test cases. 
For each test cases, the first line is one integer N (1<=N<=1000) which is the size of the image. Then there are N lines and each line has N integers each of which is either 0 or 1.

 

Output

For each test case, please output a line which is "Case X: Y", X means the number of the test case and Y means the number of the squares we are interested in in the image.

 

Sample Input

   
   
   
   

    
    
    
    1
3
1 1 0
1 1 0
0 0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 5
   
   
   
   

 

Source

2012 Multi-University Training Contest 4

 

Recommend

zhoujiaqi2010

正归的做法应该是线段树，没想到用暴力过了，呵呵

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAXN 1005
int up[MAXN][MAXN];
int lleft[MAXN][MAXN];
int map[MAXN][MAXN];
int fmin(int a,int b)
{
    if(a<b)
    return a;
    return b;
}
int main()
{
    int n,tcase,i,j,t=1;
    int ans;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&n);
        ans=0;
        memset(lleft,0,sizeof(lleft));
        memset(up,0,sizeof(up));
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&map[i][j]);
                if(map[i][j])
                {
                    ans++;
                    lleft[i][j]=lleft[i][j-1]+1;
                    up[i][j]=up[i-1][j]+1;
                }
                else
                {
                    lleft[i][j]=0;
                    up[i][j]=0;
                }
            }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
              
                int k=fmin(up[i][j],lleft[i][j]);
                if(k<=1)
                continue;
                int temp;
                for(temp=2;temp<=k;temp++)
                {
                    if(up[i][j-temp+1]>=temp&&lleft[i-temp+1][j]>=temp)
                    {
                        ans++;
                   
                    }
                }

            }
          
        }

        printf("Case %d: %d\n",t++,ans);
    }

    return 0;
}