POJ——T3352 Road Construction

http://poj.org/problem?id=3352

vis表示访问的次序 low的值相同的点在同一连通分量

 #include <algorithm>
 #include <cstring>
 #include <cstdio>

 using namespace std;

 const int N();
 int n,m,u,v,ans;
 int head[N],sumedge;
 struct Edge
 {
     int to,next;
     Edge(int to=,int next=) :
         to(to),next(next) {}
 }edge[N<<];

 void ins(int from,int to)
 {
     edge[++sumedge]=Edge(to,head[from]);
     head[from]=sumedge;
 }

 int tim,dfn[N],low[N],vis[N],du[N];
 int sumcol,col[N],colvis[N];

 void DFS(int now,int pre)
 {
     low[now]=dfn[now]=++tim;
     vis[now]=;
     for(int i=head[now];i;i=edge[i].next)
     {
         int go=edge[i].to;
         if(go==pre) continue;
         if(!vis[go])
         {
             DFS(go,now);
             low[now]=min(low[now],low[go]);
         } else if(vis[go]==)
             low[now]=min(low[now],dfn[go]);
     }
     vis[now]=;
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     for(;m;m--)
     {
         scanf("%d%d",&u,&v);
         ins(u,v); ins(v,u);
     }
     for(int i=;i<=n;i++)
         if(!vis[i]) DFS(i,i);
     for(u=;u<=n;u++)
         for(int j=head[u];j;j=edge[j].next)
         {
             v=edge[j].to;
             if(low[u]!=low[v])
                 du[low[u]]++;
         }
     for(int i=;i<=n;i++)
         if(du[i]==) ans++;
     printf("%d\n",(ans+)>>);
     return ;
 }