HDU 3926 Hand in Hand（判断同构）

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本文介绍了一种通过比较两个图的结构来判断它们是否同构的方法。具体实现了标记环或链，并按点数目排序，若点数相同则环排前。通过逐个比较节点数和环状态来判断两图是否完全相同。

Hand in Hand

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 122768/62768 K (Java/Others)
Total Submission(s): 1880 Accepted Submission(s): 635

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

Sample Output

  
   Case #1: YES
Case #2: NO

Source

2011 Multi-University Training Contest 9 - Host by BJTU

因为每个人小朋友只有两只手，所以每个点最多只有2度。图有可能是环、链，以及环和链构成的复杂图。　　

　　如何判断两幅图是否相似呢？判断相似是判断两幅图的圈的数量，以及构成圈的点数是否相同。还有判断链的数目和构成链的点数是否相同。

　　具体实现：标记环（或者链），按照点的数目排序。如果点数相同，环排在前面。然后逐个判断，如果全部都相同，就是同构。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10010;
struct node
{
    int cnt;
    bool iscc;
}arr1[N],arr2[N];
int f[N],r[N],iscir[N];
bool cmp(const node &x, const node &y)
{
    if(x.cnt!=y.cnt) return x.cnt<y.cnt;
    if(x.iscc<y.iscc) return 1;
    return 0;
}
int Find(int x)
{
    if(x==f[x]) return x;
    return f[x]=Find(f[x]);
}
void Link(int i,int j)
{
    int a=Find(i),b=Find(j);
    if(a!=b) {f[b]=a;r[a]+=r[b];}
    else iscir[a]=1;
}
void init()
{
    for(int i=0;i<N;i++)
    {
        f[i]=i;iscir[i]=0;r[i]=1;
    }
}
int main()
{
    //freopen("test.txt","r",stdin);
    int ca,k,i,j,x,y,n1,m2,n2,m1,a,b;
    k=1;
    scanf("%d",&ca);
    while(ca--)
    {
        scanf("%d%d",&n1,&m1);
        init();
        while(m1--)
        {
            scanf("%d%d",&x,&y);
            Link(x,y);
        }
        a=0;
        for(i=1;i<=n1;i++)
        {
            if(i==Find(i))
            {
                arr1[a].cnt=r[i];
                arr1[a++].iscc=iscir[i];
            }
        }
        scanf("%d%d",&n2,&m2);
        init();
        while(m2--)
        {
            scanf("%d%d",&x,&y);
            Link(x,y);
        }
        b=0;
        for(i=1;i<=n1;i++)
        {
            if(i==Find(i))
            {
                arr2[b].cnt=r[i];
                arr2[b++].iscc=iscir[i];
            }
        }
        printf("Case #%d: ",k++);
        if(n2!=n1||m2!=m1)
        {
            printf("NO\n");
            continue;
        }
        if(a==b)
        {
            sort(arr1,arr1+a,cmp);
            sort(arr2,arr2+a,cmp);
            for(i=0;i<a;i++)
            {
                if(arr1[i].cnt!=arr2[i].cnt) break;
                if(arr1[i].iscc!=arr2[i].iscc) break;
            }
            if(i==a) printf("YES\n");
            else printf("NO\n");
        }
        else printf("NO\n");
    }
    return 0;
}