poj Going from u to v or from v to u?

Going from u to v or from v to u?

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn‘t know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

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

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write ‘Yes‘ if the cave has the property stated above, or ‘No‘ otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb

思路：Tarjan+拓扑排序。

喜闻乐见的Code

#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 1010
#define maxm 6010
struct Edge {
	int u,v,next;
	Edge(int u=0,int v=0,int next=0):
		u(u),v(v),next(next) {}
}edge[maxm],EDGE[maxm];
int head[maxn],HEAD[maxn],rd[maxn];
int dfn[maxn],low[maxn],st[maxn],col[maxn];
int cnt,CNT,timee,top,numcolor;
bool ins[maxn],vis[maxn];
inline int input() {
	char c=getchar();int x=0,flag=1;
	for(;c<‘0‘||c>‘9‘;c=getchar())
		if(c==‘-‘) flag=-1;
	for(;c>=‘0‘&&c<=‘9‘;c=getchar())
		x=(x<<1)+(x<<3)+c-‘0‘;
	return x*flag;
}
inline void Add_edge(int u,int v,bool flag) {
	if(flag) edge[++cnt]=Edge(u,v,head[u]),head[u]=cnt;
	else EDGE[++CNT]=Edge(u,v,HEAD[u]),HEAD[u]=CNT;
	return;
}
void dfs(int now) {
	dfn[now]=low[now]=++timee;
	st[++top]=now;
	ins[now]=vis[now]=true;
	for(int i=head[now];i;i=edge[i].next) {
		int v=edge[i].v;
 		if(ins[v]) low[now]=min(low[now],dfn[v]);
		else if(!vis[v]) {
			dfs(v);
			low[now]=min(low[now],low[v]);
		}
	}
	if(dfn[now]==low[now]) {
		col[now]=++numcolor;
		while(st[top]!=now) {
			col[st[top]]=numcolor;
			ins[st[top--]]=false;
		}
		ins[now]=false;
		top--;
	}
	return;
}
inline bool topo() {
	int pos,count=0;
	for(int i=1;i<=numcolor;i++)
		if(rd[i]==0) pos=i,count++;
	if(count>1) return false;
	for(int k=1;k<=numcolor;k++) {
		count=0;
		for(int i=HEAD[pos];i;i=EDGE[i].next) {
			int v=EDGE[i].v;
			rd[v]--;
			if(rd[v]==0) pos=i,count++;
		}
		if(count>1) return false;
	}
	return true;
}
int main() {
	for(int t=input();t;t--) {
		int n=input(),m=input();
		for(int i=1;i<=n;i++) {
			ins[i]=vis[i]=false;
			head[i]=HEAD[i]=rd[i]=0;
			col[i]=dfn[i]=low[i]=st[i]=0;
		}
		cnt=CNT=top=timee=numcolor=0;
		for(int i=1;i<=m;i++) {
			int u=input(),v=input();
			Add_edge(u,v,1);
		}
		for(int i=1;i<=n;i++)
			if(!vis[i]) dfs(i);
		for(int i=1;i<=m;i++) {
			int u=edge[i].u,v=edge[i].v;
			if(col[u]!=col[v]) {
				Add_edge(col[u],col[v],0);
				rd[col[v]]++;
			}
		}
		topo()?printf("Yes\n"):printf("No\n");
	}
	return 0;
}