POJ 2762 Going from u to v or from v to u? Tarjan缩点+判断链

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Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13438		Accepted: 3507

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

在一个图中让你判断是否任意两个点u和v，都有u能够到达v或者v能够到达u。在强联通分量里面的点肯定满足此条件，所以首先tarjan缩点，缩点之后，重新建图，要实现u到v或者v到u这一条件，那么建的新图必须满足是一条链，如果不是链，而在u点有分支，一边连接v1，一边连接v2的话，那么v1不可能到达v2，v2也不可能到达v1，因此就不能满足条件，所以不成立。

//4616K	547MS
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define M 1007
using namespace std;
int dfn[M],low[M],head[M],vis[M],stack[M],belong[M];
int n,m,cnt,scnt,begin,num;
int g[M][M],in[M];
struct E
{
int v,to;
}edg[M*M];
void init()
{
memset(head,-1,sizeof(head));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(belong ,0,sizeof(belong));
memset(stack,0,sizeof(stack));
memset(vis,0,sizeof(vis));
memset(g,0,sizeof(g));
memset(in,0,sizeof(in));
cnt=scnt=num=begin=0;
}
void addedge(int u,int v)
{
edg[num].v=v;edg[num].to=head[u];
head[u]=num++;
}
void tarjan(int x)
{
int v;
dfn[x]=low[x]=++cnt;
stack[++begin]=x;
for(int i=head[x];i!=-1;i=edg[i].to)
{
v=edg[i].v;
if(!dfn[v])
{
tarjan(v);
low[x]=min(low[x],low[v]);
}
else if(!vis[v])
low[x]=min(low[x],dfn[v]);
}
if(low[x]==dfn[x])
{
scnt++;
do
{
v=stack[begin--];
belong[v]=scnt;
vis[v]=1;
}while(v!=x);
}
}
bool link()//判断能否形成一条链
{
queue<int>q;
for(int i=1;i<=scnt;i++)
if(!in[i])q.push(i);
if(q.size()>1)return false;
while(!q.empty())
{
int now=q.front();
q.pop();
for(int i=1;i<=scnt;i++)
if(g[now][i])
{
in[i]--;
if(!in[i])q.push(i);
}
if(q.size()>1)return false;
}
return true;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
init();
scanf("%d%d",&n,&m);
int a,b;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
addedge(a,b);
}
for(int i=1;i<=n;i++)
if(!dfn[i])tarjan(i);
if(scnt==1) {printf("Yes\n");continue;}
for(int u=1;u<=n;u++)//缩点之后，重新建图
for(int j=head[u];j!=-1;j=edg[j].to)
{
int v=edg[j].v;
if(u!=v&&belong[u]!=belong[v])
{
g[belong[u]][belong[v]]=1;
in[belong[v]]++;
}
}
if(link())printf("Yes\n");
else printf("No\n");
}
return 0;
}