hdu 3926 Hand in Hand 判断两个度最多是2的无向图是否同构

Problem Description

In order to get rid of Conan, Kaitou KID disguises himself as a teacher in the kindergarten. He knows kids love games and works out a new game called "hand in hand".

Initially kids run on the playground randomly. When Kid says "stop", kids catch others' hands immediately. One hand can catch any other hand randomly. It's weird to have more than two hands get together so one hand grabs at most one other hand. After kids stop moving they form a graph.

Everybody takes a look at the graph and repeat the above steps again to form another graph. Now Kid has a question for his kids: "Are the two graph isomorphism?"

 

 

Input

The first line contains a single positive integer T( T <= 100 ), indicating the number of datasets.
There are two graphs in each case, for each graph:
first line contains N( 1 <= N <= 10^4 ) and M indicating the number of kids and connections.
the next M lines each have two integers u and v indicating kid u and v are "hand in hand".
You can assume each kid only has two hands.

 

 

Output

For each test case: output the case number as shown and "YES" if the two graph are isomorphism or "NO" otherwise.

 

 

Sample Input

  
  

   
   2

3 2
1 2
2 3
3 2
3 2
2 1

3 3
1 2
2 3
3 1
3 1
1 2
  
  

 

 

Sample Output

  
  

   
   Case #1: YES
Case #2: NO
  
  

 

//因为度最多是2，所以每一块是链或者是一个环。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100000;
struct edge
{
    int v;
    int next;
};
int V,E;
int p[maxn];
edge G[maxn];
int l;
void init()
{
    l=0;
    memset(p,-1,sizeof(p));
}
void addedge_double(int u,int v)
{
    G[l].v=v;
    G[l].next=p[u];
    p[u]=l++;
    G[l].v=u;
    G[l].next=p[v];
    p[v]=l++;
}
int a[maxn],b[maxn];
int la,lb;
void input()
{
    init();
    scanf("%d%d",&V,&E);
    for(int i=0;i<E;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        addedge_double(u,v);
    }
}
int vis[maxn];
int cnt;
int circle;
void dfs(int u,int fath)
{
    vis[u]=1;cnt++;
    for(int i=p[u];i!=-1;i=G[i].next)
    {
        int t=G[i].v;
        if(t==fath) continue;
        if(vis[t]) circle=1;
        else dfs(t,u);
    }
}
void solve(int a[maxn],int &l)
{
    l=0;
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=V;i++)
    {
        if(!vis[i])
        {
            cnt=0;
            circle=0;
            dfs(i,-1);
            if(circle) cnt=-cnt;
            a[l++]=cnt;
        }
    }
}
int same()
{
    sort(a,a+la);
    sort(b,b+lb);
    if(la!=lb) return 0;
    for(int i=0;i<la;i++) if(a[i]!=b[i]) return 0;
    return 1;
}
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        input();
        solve(a,la);
        input();
        solve(b,lb);
        if(same()) printf("Case #%d: YES\n",pl++);
        else printf("Case #%d: NO\n",pl++);
    }
    return 0;
}