HDU 3926 Hand in Hand (图的同构)

题目链接

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

PS:这是一个图的判断的题,首先这个图的最大度是2,就说明图中可能有链,环。现在可以遍历每个连通分量,看节点数以及该连通分量是环还是链。判断之前先将两个图按照节点数排序,若节点数一样,则按照链在前环在后的顺序排序。然后遍历比较就可以了,细节看代码。

#include 
#include
#include
#include
#include
#include
#include
#include
#include
#include
const int maxn=1e4+10;
const int mod=1e9+7;
const int inf=1e8;
#define me(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
typedef long long ll;
using namespace std;
int roota[maxn],rootb[maxn];
int na,ma,nb,mb;
struct node
{
    int cnt;
    int type;
    node(int a=1,int b=0):cnt(a),type(b){}
    bool friend operator<(node a,node b)//首先按照节点数目排序,若节点数目一样,则按照链在前,环在后的顺序排列。
    {
        if(a.cnt==b.cnt)
            return a.typea[y1].cnt)
        {
            a[x1].cnt+=a[y1].cnt;
            Fa[y1]=x1;
        }
        else
        {
            a[y1].cnt+=a[x1].cnt;
            Fa[x1]=y1;
        }
    }
}
bool check()//比较函数
{
    sort(ta,ta+na+1);
    sort(tb,tb+nb+1);
    for(int i=0;i<=na;i++)
        if(ta[i].cnt!=tb[i].cnt||ta[i].type!=tb[i].type)
            return 0;
    return 1;
}
int main()
{
    int t,Case=1;
    scanf("%d",&t);
    while(t--)
    {
        inct();
        scanf("%d%d",&na,&ma);
        while(ma--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            unite(u,v,roota,ta);
        }
        scanf("%d%d",&nb,&mb);
        while(mb--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            unite(u,v,rootb,tb);
        }
        printf("Case #%d: ",Case++);
        if(ma!=mb||na!=nb)//当节点数不一样,或者手拉手的数目不一样,图肯定不一样。
            printf("NO\n");
        else if(check())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

 

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