hdu 3826 Hand in Hand 同构图★

Hand in Hand

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 122768/62768 K (Java/Others)
Total Submission(s): 2402    Accepted Submission(s): 821

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

23 21 22 33 23 22 13 31 22 33 13 11 2

 

Sample Output

Case #1: YESCase #2: NO
题目挺简单。。
细节太多。。
同构图 ：
1。点和边的个数相等。。
2.点的入度相同
3.环相同，连通分量相同
#include<stdio.h>#include<string.h>#include<algorithm>#define MAXN 100003using namespace std;struct Graph{    int son;    bool ring;} a[100003],b[100003];int pre1[100003],pre2[100003];void init(){    for(int i=0; i<100002; i++)    {        pre1[i]=i;        pre2[i]=i;        a[i].son=1;        b[i].son=1;        a[i].ring=false;        b[i].ring=false;    }}int cmp(Graph a,Graph b){    if(a.son==b.son)        return a.ring<b.ring;    return a.son<b.son;}int Find(int x, int pre[]) //查找根结点+路径压缩{    return x == pre[x] ? x : pre[x]=Find(pre[x], pre);}void Merge(int x,int y,int pre[],Graph tmp[]){    int X=Find(x,pre);    int Y=Find(y,pre);    if(X==Y)        tmp[X].ring=true;    else    {        if(tmp[X].son>=tmp[Y].son)            pre[Y]=X,tmp[X].son+=tmp[Y].son;        else            pre[X]=Y,tmp[Y].son+=tmp[X].son;    }}bool judge(int num,Graph a[],Graph b[]){    sort(a+1,a+num+1,cmp);    sort(b+1,b+num+1,cmp);    for(int i=1;i<=num;i++)    {        if(a[i].son!=b[i].son||a[i].ring!=b[i].ring)            return false;    }    return true;}int main(){    int iCase=0,T;    scanf("%d",&T);    while(T--)    {        int flag=true;        init();        int num1,match1,num2,match2;        scanf("%d%d",&num1,&match1);                for(int i=1;i<=match1;i++)        {            int x,y;            scanf("%d%d",&x,&y);            Merge(x,y,pre1,a);        }        scanf("%d%d",&num2,&match2);        if(match1!=match2)            flag=false;        for(int i=1;i<=match2;i++)        {            int x,y;            scanf("%d%d",&x,&y);            if(flag) Merge(x,y,pre2,b);        }        if(flag)        {            flag=judge(num2,a,b);             if(flag)                printf("Case #%d: YES\n", ++iCase);            else                printf("Case #%d: NO\n", ++iCase);        }        else  printf("Case #%d: NO\n", ++iCase);    }}