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  • Ice_cream's world I

    Problem Description
    ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.
     

    Input
    In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.
     

    Output
    Output the maximum number of ACMers who will be awarded.
    One answer one line.
     

    Sample Input
    8 10 0 1 1 2 1 3 2 4 3 4 0 5 5 6 6 7 3 6 4 7
     

    Sample Output
    3
     
           很简单的题     查找环的个数


    #include<stdio.h>
    #include<cstring>
    int pre[1001];
    int find(int p)  //路径压缩 
    {  
        int r=p;  
        int t;  
        while(p!=pre[p])  
        p=pre[p];  
        while(r!=p)  
        {  
            t=pre[r];  
            pre[r]=p;  
            r=t;  
        }  
        return p;  
    } 
    void merge(int x,int y)
    {
    	int fx=find(x);
    	int fy=find(y);
    	if(fx!=fy)
    	{
    		pre[fx]=fy;
    	}
    }
    int main()
    {
    	int m,n,a,b,i,flog,k,f;	
    		while(scanf("%d%d",&a,&b)!=EOF)
    		{
    			for(int i=0;i<a;i++)
    		    {
    			pre[i]=i;
    		     }
    	        flog=0;
    	       while(b--)
    		   {
    		   	scanf("%d%d",&n,&m);
    	         if(find(n)==find(m))//判断是否成环 
    	         flog++;
    	         merge(n,m);
    			} 
    		   printf("%d
    ",flog);
    		}
    	return 0;
     } 
    


    编程五分钟,调试两小时...
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  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027106.html
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