POJ2492：A Bug's Life（种类并查集）

A Bug's Life

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 45757		Accepted: 14757

题目链接：http://poj.org/problem?id=2492

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

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：

给出m对关系，他们是互斥的，问给出的关系是否有矛盾。

题解：

种类并查集，可以用带权并查集解决，这里我就多开了两个集合。

思路可以参见我之前的一篇博客：https://www.cnblogs.com/heyuhhh/p/9980330.html

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int N = 4005;
int t,n,m;
int f[N];

int find(int x){
    return f[x]==x ? f[x] :f[x]=find(f[x]); 
}
bool same(int x,int y){
    return find(x)==find(y);
}
void Union(int x,int y){
    int fx=find(x),fy=find(y);
    f[fx]=fy;
}
int main(){
    scanf("%d",&t);int tot=0;
    while(t--){
        tot++;
        bool flag=true;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=N-5;i++) f[i]=i;
        for(int i=1,x,y;i<=m;i++){
            scanf("%d%d",&x,&y);
            if(!flag) continue ;
            int fx=find(x),fy=find(y);
            if(!same(x,y) && !same(x+n,y+n)){
                Union(x,y+n);Union(x+n,y);
            }else flag=false;
        }
        printf("Scenario #%d:
",tot);
        if(flag) puts("No suspicious bugs found!");
        else puts("Suspicious bugs found!");
        printf("
");
    }
    return 0;
}