POJ2492 A Bug's Life

 

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 33833		Accepted: 11078

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.

Source

TUD Programming Contest 2005, Darmstadt, Germany

 

带偏移量的并查集。

两个集合并时，要注意处理dis数组，使点之间的关系不出错

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int mxn=20000;
int fa[mxn];
int dis[mxn];
int n,m;
bool flag;
int find(int x){
    if(fa[x]==x){
        return x;
    }
    int newfx=find(fa[x]);
    dis[x]=(dis[x]+dis[fa[x]])%2;
    return fa[x]=newfx;
}
bool pd(int a,int b){
    int u=find(a),v=find(b);
    if(u==v){
        if(dis[a]!=(dis[b])^1){
            flag=1;
            return 0;
        }
    }
    else{
        fa[u]=v;
        dis[u]=((dis[a]-dis[b])^1);
    }
    return 1;
}
void init(int x){
    for(int i=1;i<=x;i++)fa[i]=i,dis[i]=0;
}
int main(){
    int T;
    scanf("%d",&T);
    for(int ro=1;ro<=T;ro++){
        printf("Scenario #%d:\n",ro);
        scanf("%d%d",&n,&m);
        init(n);
        int i,j;
        int x,y;
        flag=0;
        for(i=1;i<=m;i++){
            scanf("%d%d",&x,&y);
            pd(x,y);
        }
        if(flag)printf("Suspicious bugs found!\n");
        else printf("No suspicious bugs found!\n");
        printf("\n");
    }
    return 0;
}