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=200010,M=2000010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=0;
    for(i=first[u];i!=-1;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=-1;
        bcnt++;
        while(x!=u){
            x=s.top();s.pop();
            bccno[x]=bcnt;
        }
    }
}

void find_bcc()
{
    int i;
    bcnt=dfs_clock=0;//mem(iscut,0);
    mem(pre,0);mem(bccno,0);
    for(i=1;i<=n;i++){
        if(!pre[i])dfs(i,-1);
    }
}

int hig;
int dfs2(int u,int p)
{
    int max1=0,max2=0;
    for (int i=first2[u];i!=-1;i=next2[i])
    {
        int v=e2[i].v;
        if (v==p) continue;
        int tmp=dfs2(v,u)+1;
        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=0;mem(first,-1);
        for(i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            adde(a,b);
        }

        find_bcc();
        mem(first2,-1);mt2=0;
        for(i=0;i<mt;i+=2){
            if(bccno[e[i].u]!=bccno[e[i].v]){
                adde2(bccno[e[i].u],bccno[e[i].v]);
            }
        }
        hig=0;
        dfs2(1,-1);

        printf("%d
",bcnt-1-hig);
    }
    return 0;
}