POJ 2528（并查集）

A Bug's Life

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 33086 Accepted: 10840

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

23 31 22 31 34 21 23 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

题意：有n个人，每个人都有一个编号，现在有m组输入（x，y），表示x喜欢y（y喜欢x），问通过这些关系能否找出同性恋？

题解：很容易想到使用并查集，但是并不是很好写呀Orz，我开始是考虑使用2个根来表示关系，突然有发现这些数字关系有可能会形成森林，这时我们需要使用数组，记录每个根的对立面的集合，每次添加到他们对立的集合，即男女集合，这样就不用考虑它是否会形成森林，直接合并即可，这里注意在一些元素根本不知道属性的情况下是不去要合并它的对立面的

#include<cstdio>  #include<cstring>  #include<cstdlib>  #include<cmath>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<set>  #include<queue>  #include<string>  #include<bitset>  #include<utility>  #include<functional>  #include<iomanip>  #include<sstream>  #include<ctime>  using namespace std;    #define N int(1e6+10)  #define inf int(0x3f3f3f3f)  #define mod int(1e9+7)  typedef long long LL;    pair<int,int>pat[N];int fa[N],Rank[N];void init(int n){for(int i=1;i<=n;i++){pat[i].first=0;Rank[i]=1;fa[i]=i;}}int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}void unio(int x,int y){x=find(x);y=find(y);if(x!=y){if(Rank[x]>Rank[y]){Rank[x]+=Rank[y];fa[y]=fa[x];}else{Rank[y]+=Rank[x];fa[x]=fa[y];}}}int main()  {  #ifdef CDZSC      freopen("i.txt", "r", stdin);      //freopen("o.txt","w",stdout);      int _time_jc = clock();  #endif  int t,n,m,cas=0,x,y;scanf("%d",&t);while(t--){int ok=1;printf("Scenario #%d:\n",++cas);scanf("%d%d",&n,&m);init(n);while(m--){scanf("%d%d",&x,&y);if(pat[x].first==0&&pat[y].first==0){pat[x].first=y;pat[y].first=x;continue;}else{int xx=find(x);int yy=find(y);if(xx==yy){ok=0;}else{if(pat[xx].first!=0){unio(y,pat[xx].first);}else{pat[xx].first=xx;}if(pat[yy].first!=0)unio(x,pat[yy].first);else{pat[yy].first=yy;}}}}puts(ok?"No suspicious bugs found!\n":"Suspicious bugs found!\n");}    return 0;  }