POJ 2492 A Bug's Life【并查集高级应用+类似食物链】

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.

 

【题意】：Hopper 在研究某种稀有虫子的性行为。他假设虫子们有两种不同的性别，而且它们只跟异性发生关系。
在他的试验里，每个虫子和它的性行为都很容易辨认，因为它们的背后印着号码。
给出一些虫子的性行为，确定是否有同性恋的虫子能推翻这个假设。某物种n只，m个描述，表示x,y为异性，判断是否有同性交配。如果发现存在同性交配输出bugs found.

【分析】：

理论上这一题和HDU-1829应该都有两种方法，一个是设置两个并查集，将冲突的两个虫子合并入两个不同的并查集中，在合并的过程中如果发现合并冲突则表明出现问题，直接记录。

第二种方法则是同时记录偏移量和合并元素，倘若出现新输入的两个元素位于同一集合且偏移量相同，表明该两个元素与同一个元素都有交集，加上他们自身也有交集，表明他们同性恋，存在bug。

【代码】：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>

using namespace std;
const int maxn = 20005;
int fa[maxn],r[maxn];
int n,m,a,b;
void init()
{
    for(int i=1;i<=n;i++)
    {
        fa[i]=i;
        r[i]=0;
    }
}

int find(int x)
{
    if(fa[x]==x) return x;
    else
    {
        int rt=find(fa[x]);
       // fa[x]=find(fa[x]);
        r[x]=r[x]^r[fa[x]];
        return fa[x]=rt;
    }
    return fa[x];
}

void join(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    fa[fx]=fy;
    r[fx]= (!(r[y]^r[x]));
    /*fa[fy]=fx;
    r[fy]=(1+r[y]+r[x])%2;*/
}

int main()
{
    int t,flag;
    int cas=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        init();//注意初始化！
        flag=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            if(find(a)!=find(b))//根不同就合并
            {
                join(a,b);
            }
            else
            {
                if(r[a]==r[b])//r[x]=r[y]可推出x和y同性，那么就说明存在bug。详见代码：
                    flag=1;
            }
            if(flag) continue;
        }

        printf("Scenario #%d:
",++cas);
        
        if(flag)
            printf("Suspicious bugs found!
");
        else
            printf("No suspicious bugs found!
");
            
        printf("
");
    }
}