POJ2492(種類並查集)

bigbigship發表於2014-06-12
並查集

題目連結：http://poj.org/problem?id=2492

A Bug's Life
Time Limit: 10000MS   Memory Limit: 65536K
Total Submissions: 27388   Accepted: 8907

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!
題目意思就是 給定n只蟲子 不同性別的可以在一起 相同性別的不能在一起
給你m對蟲子 判斷中間有沒有同性別在一起的；
我們把同性的放到一個集合裡 如果一個集合裡出現了異性 則說明存在同性戀在一起
假設 x 為一種性別 x+n為與其相反的性別  
若a,b為同性 的  我們則可以把判斷 (a，b+n)  (b,a+n)為異性反之亦然；
程式碼如下：
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 5000;
int par[maxn];
void init(int n){
   for(int i=0;i<=n;i++)
      par[i]=i;
}
int find(int x){
    if(x!=par[x])
        par[x]=find(par[x]);
    return par[x];
}
void Union(int a,int b){
    int x=find(a);
    int y=find(b);
    if(x!=y)
       par[x]=y;
}
bool judge(int x,int y){//判斷是否為同性，異性為真，同性為假
   x=find(x);
   y=find(y);
   if(x!=y)
    return true;
   return false;
}
int main()
{
    int t,n,m,cas=1;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        scanf("%d%d",&n,&m);
        init(n*2);
        bool l=1;
        int x,y;
        while(m--){
            scanf("%d%d",&x,&y);
            if(judge(x,y)||judge(x+n,y+n)){
                Union(x,y+n);//把同性的加到一個集合裡
                Union(x+n,y);
            }
            else
                l=0;
        }
        if(i!=1)
            puts("");
        printf("Scenario #%d:\n",cas++);
        if(l)
            printf("No suspicious bugs found!\n");
        else
            printf("Suspicious bugs found!\n");
    }
    return 0;
}


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