本文主要是介绍HDU 3926 Hand in Hand（同构图），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
链接：
http://acm.hdu.edu.cn/showproblem.php?pid=3926
原题：
Hand in Hand
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 122768/62768 K (Java/Others)Total Submission(s): 731 Accepted Submission(s): 294
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?"
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.
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
23 2 1 2 2 3 3 2 3 2 2 13 3 1 2 2 3 3 1 3 1 1 2
Sample Output
Case #1: YES Case #2: NO
Source
2011 Multi-University Training Contest 9 - Host by BJTU
分析与总结：
由于每个人只有两只手，每只手不能与多只手连，就是说每个节点的度数最多只能是2，最终这些学生可能会被分成多组，每组是一个环形圆圈，或者是一条链。
题目要判断两个已经连好的图是否是"同构图", 这里所谓的同构图是指两个图组成的不同的圆圈，链条，他们各个所对应的数量都是相等的。
那么我们只需要用并查集把连这的都合并起来，如果发现有环，就进行标识。然后把两个图的并查集按照每颗树的节点个数大小排序，如果个数相同，那么有环的都放在前面。 然后，只要一一比较两个已排序的数组，只要发现有一个是不同的，就不是同构图。
代码：
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 10010
using namespace std;int f[N], rank[N], n1, m1, n2, m2, pos1, pos2, isCircle[N];struct Node{int num;int isCircle;friend bool operator <(const Node&a,const Node&b){if(a.num!=b.num) return a.num<b.num;if(a.isCircle<b.isCircle) return true;return false;}
}arr1[N], arr2[N];void init(int n){memset(isCircle, 0, sizeof(isCircle));for(int i=0; i<=n; ++i)f[i]=i,rank[i]=1;
}
int find(int x){int i, j=x;while(j!=f[j]) j=f[j];while(x!=j){ i=f[x]; f[x]=j; x=i; }return j;
}
void Union(int x,int y){int a=find(x), b=find(y);if(a==b){isCircle[a] = 1;return;}rank[a] += rank[b];f[b] = a;
}
void play(int n, int m){int a,b;init(n);for(int i=0; i<m; ++i){scanf("%d%d",&a,&b);Union(a, b);}
}bool judge(){if(pos1!=pos2)return false;sort(arr1, arr1+pos1);sort(arr2, arr2+pos2);for(int i=0; i<pos1; ++i){if(arr1[i].num!=arr2[i].num)return false;if(arr1[i].isCircle!=arr2[i].isCircle)return false;}return true;
}int main(){int T,n1,cas=1;scanf("%d",&T);while(T--){printf("Case #%d: ",cas++);scanf("%d%d",&n1,&m1);play(n1,m1);pos1 = 0;for(int i=1; i<=n1; ++i)if(i==find(i)){arr1[pos1].num = rank[i];arr1[pos1++].isCircle = isCircle[i];}scanf("%d%d",&n2,&m2);play(n2, m2);if(n1!=n2 || m1!=m2){puts("NO");continue;} pos2 = 0;for(int i=1; i<=n2; ++i)if(i==find(i)){arr2[pos2].num = rank[i];arr2[pos2++].isCircle = isCircle[i];}if(judge()) puts("YES");else puts("NO");}return 0;
}—— 生命的意义，在于赋予它意义。
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