A Bug's Life

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

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

TUD Programming Contest 2005, Darmstadt, Germany

首先，初始化的時候所有節點的父節點都是自己，sex[]都為0。又因為合併時無需路徑壓縮，所以根節點的gender始終為0，這是推導其他子節點關係的關鍵。

　　Find：由根節點始終為0，我們可以得到在Findt時當前節點 i 與根節點的更新公式:sex[i] = sex[i] ^ sex[ f[i] ] （^為異或）. 即 i 和 f[i] 性別相同 則 i 和根節點是異性，否則 i 和根節點則為同性。

　　合併操作：合併x, y，找到x, y的父節點a, b，合併f[a] = b， 作為 子節點的那個父節點對根節點的關係 由sex[x] 和 sex[y]決定， 由於多加了一條邊，所以sex[a] = (sex[x] + sex[y] +1)%2.（兩個子節點的關係決定兩個父節點的關係）

#include <cstdio> 
#include <iostream>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int f[2002],sex[2002];
int find(int x){
	if(f[x]==x)
		return x;
	int t=find(f[x]); 
	sex[x]=sex[x]^sex[f[x]];
	return f[x]=t;
}

bool merge(int x,int y){
	int a=find(x);
	int b=find(y);
	if(a==b){
		if(sex[x]==sex[y])	return true;
		return false;
	}
        f[a]=b; 
    //更新被歸併的根節點的性別。
	sex[a]=(sex[x]+sex[y]+1)%2;
    return false;
}

int main(int argc, char** argv) {
	int cnt;
	scanf("%d",&cnt);
	for(int i=1;i<=cnt;i++){
		int n,m;
		scanf("%d %d",&n,&m);
		for(int j=1;j<=n;j++){
			f[j]=j;
			sex[j]=0;
		}
		int a,b;
			bool flag=false;
		 while(m--){
		 	scanf("%d %d",&a,&b);
		 	//有同一個祖先。 
		 	if(flag)continue;
		 	flag=merge(a,b);
		 }
		 if(flag)
		 	printf("Scenario #%d:\nSuspicious bugs found!\n\n",i);
		 else
		 	printf("Scenario #%d:\nNo suspicious bugs found!\n\n",i);	 
	}
	return 0;
}