poj 3352 Road Construction(边双连通分量+缩点)

题目链接：http://poj.org/problem?id=3352

这题和poj 3177 一样，参考http://www.cnblogs.com/frog112111/p/3367039.html

AC代码：

 #include<cstdio>
 #include<cstring>
 const int N=+;
 const int M=+;

 struct EDGE
 {
     int v,next;
 }edge[M*];
 int first[N],low[N],dfn[N],belong[N],degree[N],sta[M],instack[M];
 int g,cnt,top,scc;
 int min(int a,int b)
 {
     return a<b?a:b;
 }
 void AddEdge(int u,int v)
 {
     edge[g].v=v;
     edge[g].next=first[u];
     first[u]=g++;
 }
 void Tarjan(int u,int fa)
 {
     int i,v;
     low[u]=dfn[u]=++cnt;
     sta[++top]=u;
     instack[u]=;
     for(i=first[u];i!=-;i=edge[i].next)
     {
         v=edge[i].v;
         if(i==(fa^))
             continue;
         if(!dfn[v])
         {
             Tarjan(v,i);
             low[u]=min(low[u],low[v]);
         }
         else if(instack[v])
             low[u]=min(low[u],dfn[v]);
     }
     if(dfn[u]==low[u])
     {
         scc++;
         while()
         {
             v=sta[top--];
             instack[v]=;
             belong[v]=scc;
             if(v==u)
                 break;
         }
     }
 }
 int main()
 {
     int n,m,u,v,i,j;
     scanf("%d%d",&n,&m);
         g=cnt=top=scc=;
         memset(first,-,sizeof(first));
         memset(low,,sizeof(low));
         memset(dfn,,sizeof(dfn));
         memset(instack,,sizeof(instack));
         memset(degree,,sizeof(degree));
         for(i=;i<m;i++)
         {
             scanf("%d%d",&u,&v);
             {
                 AddEdge(u,v);
                 AddEdge(v,u);
             }
         }
         for(i=;i<=n;i++)
             if(!dfn[i])
                 Tarjan(,-);
         for(i=;i<=n;i++)
         {
             for(j=first[i];j!=-;j=edge[j].next)
             {
                 v=edge[j].v;
                 if(belong[i]!=belong[v])
                     degree[belong[i]]++;
             }
         }
         int sum=;
         for(i=;i<=n;i++)
             if(degree[i]==)
                 sum++;
         int ans=(sum+)/;
         printf("%d\n",ans);
     return ;
 }