POJ 2762 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

卡在读题。。。

题意：

给定一副有向图，选择两个点v和u，要求v能到达u或者u能到达v。

问是否可以对于图中的每一个点对<v, u>都能满足条件？输出Yes或No。

首先，想到tarjan缩点，之后原图变成DAG，通过观察发现只有当他是一条链时，才成立，dfs一遍就好了。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<stack>
using namespace std;
const int N=10005;
const int M=60005;
int T,n,m,cnt,tim,dcnt,hd
,low
,dfn
,belong
,x[M],y[M],du
;
bool flg,ins
;
stack<int>stk;
struct edge
{
int to,nxt;
}v[M];
void addedge(int x,int y)
{
v[++cnt].to=y;
v[cnt].nxt=hd[x];
hd[x]=cnt;
}
void tarjan(int u)
{
dfn[u]=low[u]=++tim;
stk.push(u);
ins[u]=1;
for(int i=hd[u];i;i=v[i].nxt)
if(!dfn[v[i].to])
{
tarjan(v[i].to);
low[u]=min(low[u],low[v[i].to]);
}
else if(ins[v[i].to])
low[u]=min(low[u],dfn[v[i].to]);
if(dfn[u]==low[u])
{
++dcnt;
while(1)
{
int t=stk.top();
stk.pop();
ins[t]=0;
belong[t]=dcnt;
if(u==t)
break;
}
}
}
void dfs(int u)
{
int sum=0;
for(int i=hd[u];i;i=v[i].nxt)
{
sum++;
dfs(v[i].to);
}
if(sum>1)
flg=1;
}
int main()
{
scanf("%d",&T);
while(T--)
{
flg=0;
cnt=dcnt=tim=0;
memset(hd,0,sizeof(hd));
memset(du,0,sizeof(du));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x[i],&y[i]);
addedge(x[i],y[i]);
}
for(int i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
cnt=0;
memset(hd,0,sizeof(hd));
for(int i=1;i<=m;i++)
if(belong[x[i]]!=belong[y[i]])
{
addedge(belong[x[i]],belong[y[i]]);
du[belong[y[i]]]++;
}
int sum=0;
for(int i=1;i<=dcnt;i++)
if(du[i]==0)
{
sum++;
dfs(i);
}
if(sum>1||flg)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}