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。。。。

  1 //STATUS:C++_AC_734MS_37312KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 #pragma comment(linker,"/STACK:102400000,102400000")

 24 using namespace std;

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef __int64 LL;

 34 typedef unsigned __int64 ULL;

 35 //const

 36 const int N=200010,M=2000010;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=100000,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-8;

 41 const double OO=1e15;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 struct Edge{

 59     int u,v;

 60 }e[M],e2[M];

 61 int first2[N],next2[M],mt2;

 62 //bool iscut[M];

 63 int first[N],next[M],pre[N],low[N],bccno[N];

 64 int n,m,mt,bcnt,dfs_clock;

 65 stack<int> s;

 66 

 67 int T;

 68 int vis[N];

 69 

 70 void adde(int a,int b)

 71 {

 72     e[mt].u=a;e[mt].v=b;

 73     next[mt]=first[a];first[a]=mt++;

 74     e[mt].u=b;e[mt].v=a;

 75     next[mt]=first[b];first[b]=mt++;

 76 }

 77 void adde2(int a,int b)

 78 {

 79     e2[mt2].u=a;e2[mt2].v=b;

 80     next2[mt2]=first2[a];first2[a]=mt2++;

 81     e2[mt2].u=b;e2[mt2].v=a;

 82     next2[mt2]=first2[b];first2[b]=mt2++;

 83 }

 84 

 85 void dfs(int u,int fa)

 86 {

 87     int i,v;

 88     pre[u]=low[u]=++dfs_clock;

 89     s.push(u);

 90     int cnt=0;

 91     for(i=first[u];i!=-1;i=next[i]){

 92         v=e[i].v;

 93         if(!pre[v]){

 94             dfs(v,u);

 95             low[u]=Min(low[u],low[v]);

 96           //  if(low[v]>pre[u])iscut[i]=true;   //存在割边

 97         }

 98         else if(fa==v){  //反向边更新

 99             if(cnt)low[u]=Min(low[u],pre[v]);

100             cnt++;

101         }

102         else low[u]=Min(low[u],pre[v]);

103     }

104     if(low[u]==pre[u]){  //充分必要条件

105         int x=-1;

106         bcnt++;

107         while(x!=u){

108             x=s.top();s.pop();

109             bccno[x]=bcnt;

110         }

111     }

112 }

113 

114 void find_bcc()

115 {

116     int i;

117     bcnt=dfs_clock=0;//mem(iscut,0);

118     mem(pre,0);mem(bccno,0);

119     for(i=1;i<=n;i++){

120         if(!pre[i])dfs(i,-1);

121     }

122 }

123 

124 int hig;

125 int dfs2(int u,int p)

126 {

127     int max1=0,max2=0;

128     for (int i=first2[u];i!=-1;i=next2[i])

129     {

130         int v=e2[i].v;

131         if (v==p) continue;

132         int tmp=dfs2(v,u)+1;

133         if (max1<tmp) max2=max1,max1=tmp;

134         else if (max2<tmp) max2=tmp;

135     }

136     hig=Max(hig,max1+max2);

137     return max1;

138 }

139 

140 int main()

141 {

142   //  freopen("in.txt","r",stdin);

143     int i,j,a,b,tot;

144     while(~scanf("%d%d",&n,&m) && (n || m))

145     {

146         mt=0;mem(first,-1);

147         for(i=0;i<m;i++){

148             scanf("%d%d",&a,&b);

149             adde(a,b);

150         }

151 

152         find_bcc();

153         mem(first2,-1);mt2=0;

154         for(i=0;i<mt;i+=2){

155             if(bccno[e[i].u]!=bccno[e[i].v]){

156                 adde2(bccno[e[i].u],bccno[e[i].v]);

157             }

158         }

159         hig=0;

160         dfs2(1,-1);

161 

162         printf("%d\n",bcnt-1-hig);

163     }

164     return 0;

165 }