POJ-3177 Redundant Paths 双连通分量

　　题目链接：http://poj.org/problem?id=3177

　　本题要求的就是最少添加多少条边可变无桥的连通图，和POJ1236差不多，（度为1的边双连通分量的个数+1）/2。

  1 //STATUS:C++_AC_47MS_12568KB

  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 using namespace std;

 24 //define

 25 #define pii pair<int,int>

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

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

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

 29 #define PI acos(-1.0)

 30 //typedef

 31 typedef __int64 LL;

 32 typedef unsigned __int64 ULL;

 33 //const

 34 const int N=5010;

 35 const int INF=0x3f3f3f3f;

 36 const int MOD=100000,STA=8000010;

 37 const LL LNF=1LL<<60;

 38 const double EPS=1e-8;

 39 const double OO=1e15;

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

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

 42 //Daily Use ...

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

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

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

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

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

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

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

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

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

 52 //End

 53 

 54 struct Edge{

 55     int u,v;

 56 }e[N*4];

 57 //bool iscut[N];

 58 bool vis[N*N/2];

 59 int first[N],next[N*4],pre[N],low[N],bccno[N],cnt[N];

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

 61 stack<int> s;

 62 

 63 void adde(int a,int b)

 64 {

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

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

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

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

 69 }

 70 

 71 void dfs(int u,int fa)

 72 {

 73     int i,v;

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

 75     s.push(u);

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

 77         v=e[i].v;

 78         if(!pre[v]){

 79             dfs(v,u);

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

 81         }

 82         else if(v!=fa && pre[v]<pre[u]){

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

 84         }

 85     }

 86     if(low[u]==pre[u]){

 87         int x=-1;

 88         bcnt++;

 89         while(x!=u){

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

 91             bccno[x]=bcnt;

 92         }

 93     }

 94 }

 95 

 96 void find_bcc()

 97 {

 98     int i;

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

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

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

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

103     }

104 }

105 

106 int main()

107 {

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

109     int i,j,a,b,ans,t;

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

111     {

112         mt=0;

113         mem(first,-1);

114         mem(vis,0);

115         while(m--){

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

117             if(a<b)swap(a,b);

118             t=a*(a-1)/2+b;

119             if(!vis[t]){

120                 vis[t]=true;

121                 adde(a,b);

122             }

123         }

124 

125         find_bcc();

126         ans=0;

127         if(bcnt>1){

128             mem(cnt,0);

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

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

131                     cnt[bccno[e[i].u]]++;

132                     cnt[bccno[e[i].v]]++;

133                 }

134             }

135             for(i=1;i<=bcnt;i++){

136                 if(cnt[i]<=1)ans++;

137             }

138         }

139 

140         printf("%d\n",(ans+1)/2);

141     }

142     return 0;

143 }