HDU 5222 Exploration

首先对于所有的无向边，我们使用并查集将两边的点并起来

若一条边未合并之前，两端的点已经处于同一个集合了，那么说明必定存在可行的环（因为这两个点处于同一个并查集集合中，那

么它们之间至少存在一条路径）

如果上一步没有判断出环，那么仅靠无向边是找不到环的

考虑到，处于同一个并查集集合中的点之间必定存在一条路径互达，因此将一个集合的点合并之后，原问题等价于在新生成的有向

图中是否有环（包括自环）。。

求一下强连通分量就好了。。。

Exploration
Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 333    Accepted Submission(s): 111



 

  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
 
 

  
   

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

   
 

 

  Sample Output
 
 

  
   

    YES
NO


    

     

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

   
 

 

  Source
 
 

  赛码"BestCoder"杯中国大学生程序设计冠军赛
 
 

   
 
 

  

 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
#define pb push_back
#define MP make_pair
#define fi  first
#define se second
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define SZ(X) ((int)(v).size())
#define For(i,l,r) for (int i = l; i <= r; ++i)
#define Cor(i,l,r) for (int i = l; i >= r; --i)
#define Fill(a,b) memset(a,b,sizeof(a))
#define READ freopen("a.in","r",stdin);freopen("a.out","w",stdout)
void read(string t)
{
    freopen((t+".in").c_str(),"r",stdin);
    freopen((t+".out").c_str(),"w",stdout);
}
const int inf = 0x3f3f3f3f;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VII;
typedef pair<int,int> pii;
template <class T>
void Max(T &a, T b) { a=max(a, b); }
#pragma comment(linker, "/STACK:102400000,102400000")

const int N = 1001000;
struct Edge
{
	int v, next;
	Edge() {}
	Edge(int a, int b)
	{
		v = a, next = b;
	}
} e[N<<2];
int head[N], mm;
int n, m, m1, m2;
void init_edge()
{
	mm = 0;
	memset(head, -1, sizeof head);
}
void add(int u, int v)
{
	e[mm] = Edge(v, head[u]);
	head[u] = mm++;
}

bool flag = false;
#include <stack>
stack<int> S;
int cur, scc;
int dfn[N*3], low[N*2], in[N];
bool ins[N];
void dfs(int u)
{
	dfn[u] = low[u] = ++ cur;
	S.push(u);
	ins[u] = 1;
	for (int i = head[u]; ~i; i = e[i].next)
	{
		int v = e[i].v;
		if (dfn[v]==0) {
			dfs(v);
			low[u] = min(low[u], low[v]);
		}
		else {
			if (ins[v])
				low[u] = min(low[u], dfn[v]);
		}
	}
	if (low[u] == dfn[u] ) {
		++ scc; int v;
		do {
			v =  S.top();  S.pop();
			ins[v] = 0;
			in[v] = scc;
		} while (v - u);
	}
}
int fa[N];
int f(int x) { return x==fa[x] ? x : fa[x] = f(fa[x]); }
int C[N];
int main()
{
	int re,ca=1; cin>>re;
	while (re--)
	{
		init_edge();
		cin >> n >> m1 >> m2;
		for (int i=1;i<=n;i++) fa[i] = i;
		flag = false;
		for (int i=0;i<m1;i++) {
			int u, v;
			scanf("%d%d", &u, &v);
			int x = f(u), y = f(v);
			if (x == y) {
				flag = true;
			}
			else fa[x] = y;
		}
		for (int i=0;i<m2;i++) {
			int u, v;
			scanf("%d%d", &u, &v);
			u = f(u); v = f(v);
			add(u, v);
			if (u==v) flag = true;
		}
		int tot = 0;
		memset(C,0,sizeof C);
		for (int i=1;i<=n;i++) {
			int t = f(i);
			if (C[t] == 0)
			{
				C[t] = 1;
				tot++;
			}
		}

		memset(dfn, 0 , sizeof dfn);
		memset(low,0,sizeof low);
		memset(ins, 0, sizeof ins);
		memset(in, 0, sizeof in);
		cur = scc = 0;
		while (!S.empty()) S.pop();
		for (int i=1; !flag &&i<=n;i++) {
			int t = f(i);
			if (dfn[t]==0) dfs(t);
		}
		if(scc < tot) flag = true;
		if (flag) puts("YES");
		else puts("NO");
	}
	return  0;
}