HDU 1829 A Bug's Life（种类并查集）

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1829

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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.

最近因为要给14级的新生讲数据结构，所以回头整理了一下并查集专题，重新做了一下种类并查集，然后挑了这道做个笔记。

这个题就是并查集的经典用法之：找同性恋。

题意非常简单：A喜欢B  B喜欢C  如此这般  然后去找这个关系网里面是否存在同性恋的BUG。（ps：在如今同性恋已经全美合法的情况下，这个题才是个BUG）

那么其实有2种思路，一种是维护一个并查集，然后去根据每个点到根节点的距离去添加点和判断性别，但是这样比较麻烦。

所以我采用了后一种比较容易理解的：维护2个并查集，与A性别相同的和与A性别不同的。对于每一条A喜欢B，就把B加到A的异性集合里面，然后把A加到B的异性集合里面。然后如果AB在同一个集合的话，BUG就产生了。

下面贴代码。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <map>#include <vector>#include <iostream>#include <algorithm>#define moo 1000000007//10^9+7#define PI acos(-1.0)using namespace std;int root[1000000+10];int sex[1000000+10];//性别标记，sex[x]表示x的异性集合中的其中一个int fi(int x){    while(x!=root[x])        x=root[x];    return x;}//并查集基础操作之找根void uni(int x,int y){    int t1=fi(x);    int t2=fi(y);    if(t1!=t2)        root[t2]=t1;}//并查集基础操作之合并int main(){    int T;    cin>>T;    int dd=T;    while(T--)    {        printf("Scenario #%d:\n",dd-T);        int n,m;        scanf("%d%d",&n,&m);        memset(sex,0,sizeof(sex));//异性集合初始化，0代表暂时没有发现异性        for(int i=1;i<=n;i++)            root[i]=i;        int x,y,flag=0;        for(int i=1;i<=m;i++)        {            scanf("%d%d",&x,&y);            if(flag==1)                continue;            if(fi(x)==fi(y))            {                flag=1;                continue;            }//x，y本来就在一个集合，说明他们已经确定同性，说明找出了bug            if(sex[x]!=0)                uni(sex[x],y);//合并他们的异性集合            if(sex[y]!=0)                uni(sex[y],x);            sex[x]=y;            sex[y]=x;        }        if(flag==0)            printf("No suspicious bugs found!\n\n");        else            printf("Suspicious bugs found!\n\n");    }    return 0;}