今天听见某大佬说POJ2186是连通图的入门题,然而我这个菜鸡什么都不会,因此决定记录下来...
首先贴大佬博客Orz:https://www.cnblogs.com/void/articles/2048209.html
https://blog.csdn.net/chang_mu/article/details/38709047
附上AC代码:
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
#define ll long long
const int MAXN=10005;
const int MAXM=50005;
int n,m;
struct Edge
{
int to,next;
}edge[MAXM];
int head[MAXN],tol;
int low[MAXN],dfn[MAXN],belong[MAXN];
stack<int>s;
int index;
int scc;
bool vis[MAXN];
int num[MAXN];
void Tarjan(int u)
{
int v;
low[u]=dfn[u]=++index;
s.push(u);
vis[u]=true;
for(int i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].to;
if(!dfn[v])
{
Tarjan(v);
if(low[u]>low[v])
low[u]=low[v];
}
else if(vis[v]&&low[u]>dfn[v])
low[u]=dfn[v];
}
if(low[u]==dfn[u])
{
scc++;
do{
v=s.top();
s.pop();
vis[v]=false;
belong[v]=scc;
num[scc]++;
}while(v!=u);
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(head,-1,sizeof(head));
memset(dfn,0,sizeof(dfn));
memset(vis,false,sizeof(vis));
memset(num,0,sizeof(num));
tol=0;
index=scc=0;
int u,v;
for(int i=0;i<m;i++)
{
scanf("%d%d",&u,&v);
edge[tol].to=v;
edge[tol].next=head[u];
head[u]=tol++;
}
for(int i=1;i<=n;i++)
if(!dfn[i])
Tarjan(i);
if(scc==1)
{
printf("%d\n",n);
continue;
}
int out[MAXN];
memset(out,0,sizeof(out));
for(int i=1;i<=n;i++)
for(int j=head[i];j!=-1;j=edge[j].next)
if(belong[i]!=belong[edge[j].to])
out[belong[i]]++;
int cnt=0,ans=0;
for(int i=1;i<=scc;i++)
if(out[i]==0)
{
cnt++;
ans=num[i];
}
if(cnt!=1)
printf("0\n");
else
printf("%d\n",ans);
}
return 0;
}