# HDOJ3926：Hand in hand——并查集、同构图应用

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 有点难度的一道题，分析过程参考D_Double同学的。

```#include
#include
#include
#include
#include
#include
using namespace std;
const int maxN = 10001;

int n, m, n2, m2;
int father[maxN], PT[maxN], isCircle[maxN];
//memset(0, isCircle, sizeof(isCircle));
struct node
{
int num, isCircle;
friend bool operator < (node a, node b)
{
if (a.num != b.num) return a.num>b.num;
return a.isCircle>b.isCircle;
}
}an[maxN], bn[maxN];

int findset(int x)
{
if (x == father[x]) return x;
return father[x] = findset(father[x]);
}
void Union(int x, int y)
{
x = findset(x); y = findset(y);
if (x == y)
{
isCircle[x] = 1;
return;
}
if (PT[x]>PT[y])
{
father[y] = x;
PT[x] += PT[y];
}
else
{
father[x] = y;
PT[y] += PT[x];
}
}

int main()
{
int t, ncase = 1;
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &n, &m);
memset(isCircle, 0, sizeof(isCircle));
for (int i = 1; i <= n; i++) father[i] = i, PT[i] = 1;
int u, v;
for (int i = 0; i