判断bug中是否有guy。。。。。。。。。。

用n和n+max表示逻辑正反两反面，两个同一集合内的两个都小于max代表有相同，一个小于一个大于代表不同



#include <iostream>

using namespace std;

const int maxn = 10010;
int pre[2*maxn];

int n,m;
int p1,p2;
int total;

int find(int x){
	int i,j,r;
	r = x;
	while(r != pre[r]){
		r = pre[r];
	}

	i = x;
	while(i != r){
		j = pre[i];
		pre[i] = r;
		i = j;
	}

	return r;
}

void join(int x , int y){
	int fx = find(x);
	int fy = find(y);
	if(fx != fy){
		pre[fx] = fy;
		--total;//不要放在外面。因为有一种输入是(1 2) (1 2),这样的话total就减重复了
	}
}


int main(){
	int t;
	scanf("%d",&t);

	int counter = 1;
	while(t--){
		int i;
		for(i = 0 ; i < 2*maxn ; ++i){
			pre[i] = i;
		}

		scanf("%d%d",&n,&m);

		bool flag = true;
		for(i = 0 ; i < m ; ++i){
			int a,b;
			scanf("%d%d",&a,&b);

			if(find(a) == find(b)){
				flag = false;
			}else{
				join(a,b+maxn);
				join(a+maxn,b);
			}
		}

		printf("Scenario #%d:\n",counter++);
//		printf(flag?"No suspicious bugs found!\n\n":"Suspicious bugs found!\n\n");

		if(flag){
			printf("No suspicious bugs found!\n\n");
		}else{
			printf("Suspicious bugs found!\n\n");
		}
	}

	return 0;
}