ZOJ 3811 Untrusted Patrol The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

Description

Edward is a rich man. He owns a large factory for health drink production. As a matter of course, there is a large warehouse in the factory.

To ensure the safety of drinks, Edward hired a security man to patrol the warehouse. The warehouse has N piles of drinks and M passageways connected them (warehouse is not big enough). When the evening comes, the security man will start to patrol the warehouse following a path to check all piles of drinks.

Unfortunately, Edward is a suspicious man, so he sets sensors on K piles of the drinks. When the security man comes to check the drinks, the sensor will record a message. Because of the memory limit, the sensors can only record for the first time of the security man's visit.

After a peaceful evening, Edward gathered all messages ordered by recording time. He wants to know whether is possible that the security man has checked all piles of drinks. Can you help him?

The security man may start to patrol at any piles of drinks. It is guaranteed that the sensors work properly. However, Edward thinks the security man may not works as expected. For example, he may digs through walls, climb over piles, use some black magic to teleport to anywhere and so on.

Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

The first line contains three integers N (1 <= N <= 100000), M (1 <= M <= 200000) and K (1 <= K <= N).

The next line contains K distinct integers indicating the indexes of piles (1-based) that have sensors installed. The following M lines, each line contains two integers A_i and B_i (1 <= A_i, B_i <= N) which indicates a bidirectional passageway connects piles A_i and B_i.

Then, there is an integer L (1 <= L <= K) indicating the number of messages gathered from all sensors. The next line contains L distinct integers. These are the indexes of piles where the messages came from (each is among the K integers above), ordered by recording time.

Output

For each test case, output "Yes" if the security man worked normally and has checked all piles of drinks, or "No" if not.

Sample Input

2
5 5 3
1 2 4
1 2
2 3
3 1
1 4
4 5
3
4 2 1
5 5 3
1 2 4
1 2
2 3
3 1
1 4
4 5
3
4 1 2

Sample Output

No
Yes

Source

The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

 

 

题意：给出n个点，m条边，在其中k个点上有传感器，有一个人在图上走。再给出长度为l的序列，为传感器第一次感知有人走到的顺序。求该人的行走是否合法且遍历全图。

思路：设没有传感器的点为普通点，有传感器的点为特殊点。先把普通点加入并查集中。然后按照l序列的顺序，把特殊点变为普通点，然后加入并查集。判断该点和之前的一个点是否联通，不联通则违法。注意还有两个要特判，l不等于k以及整张图不是联通图，这两种情况都要输出no。

 /*
  * Author:  Joshua
  * Created Time:  2014年09月09日 星期二 17时27分53秒
  * File Name: zoj3811.cpp
  */
 #include<cstdio>
 #include<vector>
 #include<cstring>
 using namespace std;

 #define maxn 100005

 typedef long long LL;
 int f[maxn],n,m,k,T;
 bool p[maxn];
 vector<int> r[maxn];

 void init()
 {
     int l,v,u,x,y;
     memset(p,,sizeof(p));
     scanf("%d%d%d",&n,&m,&k);
     for (int i=;i<=k;++i)
     {
         scanf("%d",&x);
         p[x]=true;
     }
     for (int i=;i<=n;++i) r[i].clear();
     for (int i=;i<=m;++i)
     {
         scanf("%d%d",&u,&v);
         r[u].push_back(v);
         r[v].push_back(u);
     }
     for (int i=;i<=n;++i) f[i]=i;
 }

 int gf(int x)
 {
     if (f[x]==x) return x;
     return (f[x]=gf(f[x]));
 }

 void update(int x)
 {
     int fx,fy;
     for (int j=;j<r[x].size();++j)
         if (!p[r[x][j]])
         {
             fx=gf(x);
             fy=gf(r[x][j]);
             if (fx<fy) f[fy]=fx;
             else f[fx]=fy;
         }
 }

 void solve()
 {
     int l,x,y;
     scanf("%d",&l);
     if (l!=k)
     {
         for (int i=;i<=l;++i)
             scanf("%d",&x);
         printf("No\n");
         return;
     }
     for (int i=;i<=n;++i)
         if (!p[i])
             update(i);
     for (int i=;i<=l;++i)
     {
         scanf("%d",&x);
         p[x]=false;
         update(x);
         if ((i>) && (gf(x)!=gf(y)))
         {
             printf("No\n");
             for (int j=i+;j<=l;++j) scanf("%d",&x);
             return;
         }
         y=x;
     }
     for (int i=;i<=n;++i)
         if (gf(i)!=)
         {
             printf("No\n");
             return;
         }
     printf("Yes\n");
 }
 int main()
 {
     scanf("%d",&T);
     while (T--)
     {
         init();
         solve();
     }
     return ;
 }