杭电 1829 A Bug's Life 经典并查集

    卧槽，写这道题写了两天啊，，，两天啊，，，容易吗。。。搞基的bug伤不起啊。。。。并查集扩展，题目：

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3921    Accepted Submission(s): 1266

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.
 

ac代码：

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
const int N=12010;
int p[N],relation[N];
int find(int x){
	int px=p[x];
	if(x==p[x])return x;
	p[x]=find(p[x]);
	relation[x]=(relation[x]+relation[px])%2;
	return p[x];
}
int main(){
	//freopen("in.txt","r",stdin);
	int numcase,k=0;
	scanf("%d",&numcase);
	while(numcase--){
	  int n,m;
	  scanf("%d%d",&n,&m);
	  int a,b,flag=0;
	  for(int i=1;i<=n;++i){
	    p[i]=i;
		relation[i]=0;
	  }
	  while(m--){
	    scanf("%d%d",&a,&b);
		if(flag==1)
			continue;
		int roota=find(a);
		int rootb=find(b);
		if(roota==rootb){
			if(relation[a]==relation[b]){
			  flag=1;
			}
		}
		else{
		  p[rootb]=roota;
		  relation[rootb]=((relation[a]+1)%2+relation[b])%2;
		}
	  }
	  printf("Scenario #%d:\n",++k);
	  if(flag)puts("Suspicious bugs found!");
	  else puts("No suspicious bugs found!");
	  puts("");
	}
	return 0;
}