zoj 3811||牡丹江网赛 c题 并查集

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5343

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

25 5 31 2 41 22 33 11 44 534 2 15 5 31 2 41 22 33 11 44 534 1 2

Sample Output

NoYes

解题思路：先把  没有  摄像头的  货物并在一起然后按照给定的顺序  来   并入有摄像头的  货物如果  中途出现   不能并入的，那么就说明   没有可能 出现给定的顺序
比如  4 2 1，我并入第二个的时候  必须要借助  1，那么就不行了，因为 1会出现在2的前面

整体  思路就是这样了，最后再判断一下   所有货物的root是否相同就可以了，如果不同的话，说明它连  联通图都构不成

#include<stdio.h>#include <string.h>#include <iostream>#include <queue>using namespace std;vector <int> v[100005];int x,f[100005],vis[100005];int n,m,l,k;void init(){    for(int i=0;i<100005;i++)        f[i]=i;}int find(int x){    if(f[x]==x)         return x;    return f[x]=find(f[x]);}void uni(int x,int y){    x=find(x);    y=find(y);    if(x!=y)        f[x]=y;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(vis,0,sizeof(vis));        memset(v,0,sizeof(v));        scanf("%d%d%d",&n,&m,&k);        int a,b;        for(int i=0;i<k;i++)        {            scanf("%d",&a);            vis[a]=1;        }        init();        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            if(vis[a]==0&&vis[b]==0)                uni(a,b);            if(vis[a])                v[a].push_back(b);            if(vis[b])                v[b].push_back(a);        }        scanf("%d%d",&l,&a);        vis[a]=0;        for(int i=0;i<v[a].size();i++)        {            if(!vis[v[a][i]])                uni(v[a][i],a);        }        int flag=1;        if(l<k)            flag=0;        for(int i=1;i<l;i++)        {            scanf("%d",&b);            vis[b]=0;            if(flag==0)continue;            for(int j=0;j<v[b].size();j++)                if(!vis[v[b][j]])                    uni(v[b][j],b);            if(find(a)!=find(b))                flag=0;        }        for(int i=2;i<=n;i++)            if(find(i)!=find(1))            {                flag=0;                break;            }        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}