poj 2762 Going from u to v or from v to u? trajan+拓扑

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
题解：求这图是不是单联通；缩点以后这图一定是一个链；
数据
3
3 2
1 2
3 2
5 4
1 2
2 3
3 4
4 5
5 4
1 2
1 3
3 4
3 5
No
Yes
No

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define inf 2000000001
int scan()
{
int res = 0 , ch ;
while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
{
if( ch == EOF )  return 1 << 30 ;
}
res = ch - '0' ;
while( ( ch = getchar() ) >= '0' && ch <= '9' )
res = res * 10 + ( ch - '0' ) ;
return res ;
}
struct is
{
int u,v;
int next;
}edge[50010],edgetop[50010];
int head[50010];
int belong[50010];
int dfn[50010];
int low[50010];
int stackk[50010];
int instack[50010];
int du[100010];
int n,m,jiedge,lu,bel,top,jiedgetop;
void update(int u,int v)
{
jiedge++;
edge[jiedge].u=u;
edge[jiedge].v=v;
edge[jiedge].next=head[u];
head[u]=jiedge;
}
void updatetop(int u,int v)
{
jiedgetop++;
edgetop[jiedgetop].u=u;
edgetop[jiedgetop].v=v;
edgetop[jiedgetop].next=head[u];
head[u]=jiedgetop;
}
void dfs(int x)
{
dfn[x]=low[x]=++lu;
stackk[++top]=x;
instack[x]=1;
for(int i=head[x];i;i=edge[i].next)
{
if(!dfn[edge[i].v])
{
dfs(edge[i].v);
low[x]=min(low[x],low[edge[i].v]);
}
else if(instack[edge[i].v])
low[x]=min(low[x],dfn[edge[i].v]);
}
int ne;
if(low[x]==dfn[x])
{
//cout<<x<<" "<<"XXX"<<endl;
bel++;
do
{
ne=stackk[top--];
belong[ne]=bel;
instack[ne]=0;
}while(x!=ne);
}
}
void tarjan()
{
memset(dfn,0,sizeof(dfn));
bel=lu=top=0;
for(int i=1;i<=n;i++)
if(!dfn[i])
dfs(i);
}
int topsort()
{
int st;
int flag=0;
for(int i=1;i<=bel;i++)
{
if(du[i]==0)
{
flag++;
st=i;
}
}
if(flag>1) return 0;
int n=bel,en;
while(n--)
{
int flagg=0;
for(int i=head[st];i;i=edgetop[i].next)
{
en=edgetop[i].v;
du[en]--;
if(du[en]==0)
{
flagg++;
st=en;
}
}
if(flagg>1) return 0;
}
return 1;
}
int main()
{
int i,t;
int nn;
scanf("%d",&nn);
while(nn--)
{
scanf("%d%d",&n,&m);
memset(head,0,sizeof(head));
memset(belong,0,sizeof(belong));
jiedge=0;
for(i=1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
update(u,v);
}
tarjan();
memset(head,0,sizeof(head));
memset(du,0,sizeof(du));
jiedgetop=0;
/* cout<<bel<<endl;
for(i=1;i<=n;i++)
cout<<belong[i]<<endl;*/
for(int i=1;i<=m;i++)
{
int u=edge[i].u;
int v=edge[i].v;
if(belong[u]!=belong[v])
{
du[belong[v]]++;
updatetop(belong[u],belong[v]);
}
}
if(topsort())
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
/*
3 2
1 2
3 2
*/

View Code