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  • hdoj-1289-A Bug's Life【种类并查集】

    A Bug's Life

    Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 11386 Accepted Submission(s): 3709


    Problem Description
    Background
    Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.

    Problem
    Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

    Input
    The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

    Output
    The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.

    Sample Input
    2 3 3 1 2 2 3 1 3 4 2 1 2 3 4

    Sample Output
    Scenario #1: Suspicious bugs found! Scenario #2: No suspicious bugs found!
    Hint
    Huge input,scanf is recommended.

    Source

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    #include<stdio.h>
    int root[2010];
    int rank[2010];
    int find_set(int x){
    	if(root[x]==x) return x;
    	int t=root[x];
    	root[x]=find_set(t);
    	rank[x]=(rank[x]+rank[t])%2;
    	return root[x];
    }
    void unio(int a,int b){
    	int  t,k;
    	t=root[a];
    	k=root[b];
    	root[t]=k;
    	rank[t]=(rank[b]+1-rank[a]);
    	return ;
    }
    int main(){
    	int t,cas=0;;
    	scanf("%d",&t);
    	while(t--){
    		printf("Scenario #%d:
    ",++cas);
    		int n,m,i,ok=0,a,b;
    		scanf("%d%d",&n,&m);
    		for(i=0;i<=n;++i){
    			root[i]=i;
    			rank[i]=0;
    		}
    		for(i=0;i<m;++i){
    			scanf("%d%d",&a,&b);  
    			if(!ok){
    				int pa,pb;
    				pa=find_set(a) ;
    				pb=find_set(b);
    				if(pa==pb&&(rank[a]==rank[b])){
    					ok=1;
    				}
    				else
    				unio(a,b);
    			}	
    		}
    		if(ok)
    		printf("Suspicious bugs found!
    
    ");
    		else
    		printf("No suspicious bugs found!
    
    ");
    		
    	}
    	return 0;
    }



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  • 原文地址:https://www.cnblogs.com/brucemengbm/p/6999224.html
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