hdu5222（拓扑排序+并查集）

Exploration

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1416    Accepted Submission(s): 388

Problem Description

Miceren likes exploration and he found a huge labyrinth underground! 

This labyrinth has N caves and some tunnels connecting some pairs of caves. 

There are two types of tunnel, one type of them can be passed in only one direction and the other can be passed in two directions. Tunnels will collapse immediately after Miceren passing them. 

Now, Miceren wants to choose a cave as his start point and visit at least one other cave, finally get back to start point. 

As his friend, you must help him to determine whether a start point satisfing his request exists.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with three integers N, M1, M2, indicating the number of caves, the number of undirectional tunnels, the number of directional tunnels. 

The next M1 lines contain the details of the undirectional tunnels. Each line contains two integers u, v meaning that there is a undirectional tunnel between u, v. (u ≠ v) 

The next M2 lines contain the details of the directional tunnels. Each line contains integers u, v meaning that there is a directional tunnel from u to v. (u ≠ v)

T is about 100.

1 ≤ N,M1,M2 ≤ 1000000.

There may be some tunnels connect the same pair of caves.

The ratio of test cases with N > 1000 is less than 5%.

 

Output

For each test queries, print the answer. If Miceren can do that, output "YES", otherwise "NO".

 

Sample Input

25 2 11 21 24 54 2 21 22 34 34 1

 

Sample Output

YESNO
Hint
If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.

此题题意：有双向桥和单向桥，走过每座桥之后该桥都会塌，问该图里面是否存在回路（就是所谓的环）

分析：

由于有双向桥，用并查集来处理双向桥的话，如果一条没有连接的边的两个点的父亲节点相同说明肯定有通路= =这是第一种情况。

然后就是剩下的单向桥，因为将双向桥用并查集连接起来了，所以接下来就相当于查找单相桥是否存在有向回路，判断有向回路的话就直接拓扑排序来查看是否能进行拓扑排序就ok！

#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>using namespace std;typedef long long ll;const int maxn=1000010,M=1000010;int head[maxn],ip,rd[maxn],f[maxn];int x[M],y[M];char c[M];struct data{    int v,next;}tu[M];void init(int n){    memset(head,-1,sizeof(head));    ip=0;    for(int i=0;i<=n;i++)///注意这里的等号！！！        f[i]=i,rd[i]=0,head[i]=-1;}void add(int u,int v){    tu[ip].v=v,tu[ip].next=head[u],head[u]=ip++;}int fa(int x){    if(f[x]!=x)        f[x]=fa(f[x]);    return f[x];}int s;void topu(int n){    queue<int>q;    for(int i=1;i<=n;i++)///注意容易错！从0开始还是从1！    {        if(!rd[i])        {q.push(i);}    }    s=0;    while(!q.empty())    {        int tem=q.front();        q.pop();        s++;        for(int i=head[tem];i!=-1;i=tu[i].next)        {            int hh=tu[i].v;            rd[hh]--;            if(!rd[hh])                q.push(hh);        }    }    //printf("%d\n",s);    if(s<n)        printf("YES\n");    else printf("NO\n");}int main(){    int t;    cin>>t;    while(t--)    {        int n,m1,m2;        cin>>n>>m1>>m2;        init(n);        int flag=0;        for(int i=0;i<m1;i++)        {            int x,y;            scanf("%d%d",&x,&y);            if(fa(x)!=fa(y))                f[fa(x)]=fa(y);            else                flag=1;        }        for(int i=0;i<m2;i++)        {            int x,y;            scanf("%d%d",&x,&y);            if(!flag)            {            x=fa(x),y=fa(y);            add(x,y);            rd[y]++;            }        }        if(flag)            printf("YES\n");        else topu(n);    }    return 0;}