HDU 3926 图的同构

Hand in Hand

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=3926

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

题意

给你两个无向图，判断是结构是否相同，每个点最多两个度。

题解

因为每个点最多只有两个度，所以只会存在链或者环，这样就容易做了，搜索就行了。特别注意１—>2 ,2->1不能看成一个环，只是一条链，和单独的１－＞２是一样的，

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 10050
#define M 100050
struct Edge{int x,y,s;}G1[M],G2[M];
int last1[N],last2[N];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
int dfs(int x,Edge*G,int *last,int *f)
{
    f[x]=1;
    int ans=1;
    for(int i=last[x];i;i=G[i].s)
        {
            Edge &e=G[i];
            if (f[e.y])continue;
            ans+=dfs(e.y,G,last,f);
        }
    return ans;
}
void init(int &n,int &m,Edge*G,int *last,int *a,int &num)
{
    static int d[N],f[N];
    int tot=0;
    num=0;
    read(n); read(m);
    memset(f,0,sizeof(int)*(n+1));
    memset(d,0,sizeof(int)*(n+1));
    memset(last,0,sizeof(int)*(n+1));
    for(int i=1;i<=m;i++)
    {
        int x,y;
        read(x); read(y);
        G[++tot]=Edge{x,y,last[x]};
        last[x]=tot;
        G[++tot]=Edge{y,x,last[y]};
        last[y]=tot;
        d[x]++; d[y]++;
       
    }
    for(int i=1;i<=n;i++)
        if (d[i]==1&&!f[i])a[++num]=dfs(i,G,last,f);
    for(int i=1;i<=n;i++)
        if (!f[i]){a[++num]=dfs(i,G,last,f);if (a[num]>2)a[num]+=N;}
}
void work()
{
    static int cas=0,n1,m1,n2,m2,a1[N],a2[N],num1,num2;
    init(n1,m1,G1,last1,a1,num1);
    init(n2,m2,G2,last2,a2,num2);
    sort(a1+1,a1+num1+1);
    sort(a2+1,a2+num2+1);
    for(int i=1;i<=num1;i++)
        if (a1[i]!=a2[i]||num1!=num2){printf("Case #%d: NO\n",++cas);return;}
    printf("Case #%d: YES\n",++cas);
        
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    int T;
    read(T);
    while(T--)work();
}