HDU-4612 Warm up 边双连通分量+缩点+最长链

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4612

　　简单图论题，先求图的边双连通分量，注意，此题有重边（admin还逗比的说没有重边），在用targan算法求的时候，处理反向边需要标记边，然后缩点，在树上求最长链。。

　　此题在比赛的时候，我的模板数组开小，WA一下午，sd。。。。

 //STATUS:C++_AC_734MS_37312KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 #pragma comment(linker,"/STACK:102400000,102400000")
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=,M=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 struct Edge{
     int u,v;
 }e[M],e2[M];
 int first2[N],next2[M],mt2;
 //bool iscut[M];
 int first[N],next[M],pre[N],low[N],bccno[N];
 int n,m,mt,bcnt,dfs_clock;
 stack<int> s;

 int T;
 int vis[N];

 void adde(int a,int b)
 {
     e[mt].u=a;e[mt].v=b;
     next[mt]=first[a];first[a]=mt++;
     e[mt].u=b;e[mt].v=a;
     next[mt]=first[b];first[b]=mt++;
 }
 void adde2(int a,int b)
 {
     e2[mt2].u=a;e2[mt2].v=b;
     next2[mt2]=first2[a];first2[a]=mt2++;
     e2[mt2].u=b;e2[mt2].v=a;
     next2[mt2]=first2[b];first2[b]=mt2++;
 }

 void dfs(int u,int fa)
 {
     int i,v;
     pre[u]=low[u]=++dfs_clock;
     s.push(u);
     int cnt=;
     for(i=first[u];i!=-;i=next[i]){
         v=e[i].v;
         if(!pre[v]){
             dfs(v,u);
             low[u]=Min(low[u],low[v]);
           //  if(low[v]>pre[u])iscut[i]=true;   //存在割边
         }
         else if(fa==v){  //反向边更新
             if(cnt)low[u]=Min(low[u],pre[v]);
             cnt++;
         }
         else low[u]=Min(low[u],pre[v]);
     }
     if(low[u]==pre[u]){  //充分必要条件
         int x=-;
         bcnt++;
         while(x!=u){
             x=s.top();s.pop();
             bccno[x]=bcnt;
         }
     }
 }

 void find_bcc()
 {
     int i;
     bcnt=dfs_clock=;//mem(iscut,0);
     mem(pre,);mem(bccno,);
     for(i=;i<=n;i++){
         if(!pre[i])dfs(i,-);
     }
 }

 int hig;
 int dfs2(int u,int p)
 {
     int max1=,max2=;
     for (int i=first2[u];i!=-;i=next2[i])
     {
         int v=e2[i].v;
         if (v==p) continue;
         int tmp=dfs2(v,u)+;
         if (max1<tmp) max2=max1,max1=tmp;
         else if (max2<tmp) max2=tmp;
     }
     hig=Max(hig,max1+max2);
     return max1;
 }

 int main()
 {
   //  freopen("in.txt","r",stdin);
     int i,j,a,b,tot;
     while(~scanf("%d%d",&n,&m) && (n || m))
     {
         mt=;mem(first,-);
         for(i=;i<m;i++){
             scanf("%d%d",&a,&b);
             adde(a,b);
         }

         find_bcc();
         mem(first2,-);mt2=;
         for(i=;i<mt;i+=){
             if(bccno[e[i].u]!=bccno[e[i].v]){
                 adde2(bccno[e[i].u],bccno[e[i].v]);
             }
         }
         hig=;
         dfs2(,-);

         printf("%d\n",bcnt--hig);
     }
     return ;
 }