POJ 2762 Going from u to v or from v to u?（强连通分量+拓扑排序）@

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

给定一个有向图，问是否任意两个点x,y，是否X可以到达Y，或Y可以到达X，一开始想成要满足环的条件，但题目要求的是或，意味着这个图只要有一个入度为零的强连通分量

就可以了，先缩点，再用拓扑排序

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const int N = 250000+10;
int cnt, taj, top, time;
int vis
, head
, stack1
, low
, dfn
, belong
;
void init()
{
memset(vis,0,sizeof(vis));
memset(head,-1,sizeof(head));
memset(low,-1,sizeof(low));
memset(dfn,-1,sizeof(dfn));
cnt=0, top=0, taj=0, time=0;
return ;
}

struct node
{
int from, to, next;
}p[250010*8];

void add(int u,int v)
{
p[cnt].from=u, p[cnt].to=v, p[cnt].next=head[u], head[u]=cnt++;
return ;
}

void dfs(int u)
{
low[u]=time;
dfn[u]=time++;
vis[u]=1, stack1[top++]=u;
for(int i=head[u];i!=-1;i=p[i].next)
{
int v=p[i].to;
if(dfn[v]==-1)
{
dfs(v);
low[u]=min(low[u], low[v]);
}
else if(vis[v])
low[u]=min(low[u], dfn[v]);
}
if(dfn[u]==low[u])
{
taj++;
while(1)
{
int now=stack1[--top];
vis[now]=0;
belong[now]=taj;
if(now==u)
break;
}
}
return ;
}
int in
, out
;
vector<int>G
;

int main()
{
int t, n, m;
scanf("%d", &t);
while(t--)
{
init();
scanf("%d %d", &n, &m);
for(int i=1;i<=m;i++)
{
int x, y;
scanf("%d %d", &x, &y);
add(x, y);
}
for(int i=1;i<=n;i++)
{
if(dfn[i]==-1)
dfs(i);
}
if(taj==1)
{
printf("Yes\n");
continue;
}
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
priority_queue<int,vector<int>, greater<int> >q;
while(!q.empty()) q.pop();
for(int i=0;i<=taj;i++) G[i].clear();
for(int i=0;i<cnt;i++)
{
int u=p[i].from, v=p[i].to;
if(belong[u]!=belong[v])
{
out[belong[u]]++, in[belong[v]]++;
G[belong[u]].push_back(belong[v]);
}
}
for(int i=1;i<=taj;i++)
{
if(in[i]==0) q.push(i);
}
if(q.size()==0)
printf("Yes\n");
else if(q.size()>1)
printf("No\n");
else
{
int flag=0;
while(!q.empty())
{
if(q.size()>1)
{
flag=2;
break;
}
int u=q.top();q.pop();
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
in[v]--;
if(in[v]==0) q.push(v);
}
}
if(flag==0)
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}