POJ 2186 强连通分量

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比较简单吧。。套模板。。

Gabow

#include "cstdlib" #include "cctype" #include "cstring" #include "cstdio" #include "cmath" #include "algorithm" #include "vector" #include "string" #include "iostream" #include "sstream" #include "set" #include "queue" #include "stack" #include "fstream" #include "strstream" using namespace std;#define  M 20000              //题目中可能的最大点数       int STACK[M],top=0;          //Gabow 算法中的辅助栈 int STACK2[M],top2=0;        // int DFN[M];                  //深度优先搜索访问次序 int ComponetNumber=0;        //有向图强连通分量个数 int Index=0;                 //索引号 int Belong[M];               //某个点属于哪个连通分支 vector Edge[M];        //邻接表表示 vector Component[M];   //获得强连通分量结果 bool bcount[M]={0};void Gabow(int i) {     int j;     DFN[i]=Index++;     STACK[++top]=i;     STACK2[++top2]=i;     for (int e=0;e<Edge[i].size();e++)     {         j=Edge[i][e];         if (DFN[j]==-1)  Gabow(j);         else if (Belong[j]==-1)       //如果访问过，但没有删除，维护STACK2         {             while(DFN[STACK2[top2]]>DFN[j])        //删除构成环的顶点                 top2--;         }     }     if(i==STACK2[top2])               //如果Stack2 的顶元素等于i，输出相应的强连通分量     {         --top2; ++ComponetNumber;         do         {             Belong[STACK[top]]=ComponetNumber;             Component[ComponetNumber].push_back(STACK[top]);         }while ( STACK[top--]!=i);     } }void solve(int N)     //此图中点的个数，注意是0-indexed！ {     memset(STACK,-1,sizeof(STACK));     memset(STACK2,-1,sizeof(STACK2));     memset(Belong,-1,sizeof(Belong));     memset(DFN,-1,sizeof(DFN));    for(int i=0;i<N;i++)         if(DFN[i]==-1)             Gabow(i);    } void reshape(int N)     //缩点形成新图,N为图中的点数 {     //bool ComDIG[M][M];     //memset(ComDIG,0,sizeof(ComDIG));     for( int i =0 ; i<  N;i++)       //一个顶点是i     {         for(int j=0;j<Edge[i].size();j++) //另外一个顶点是edge[i][j]         {             if(Belong[i]!=Belong[Edge[i][j]])                 bcount[Belong[i]]=true;             //    ComDIG[Belong[i]][Belong[Edge[i][j]]]=true;         }     }     /*for( int i =1 ;i<=ComponetNumber;i++)     {         for(int j=1;j<=ComponetNumber;j++)         {             if(ComDIG[i][j])                 cout<<i<<"     "<<j<<endl;         }     }*/} /* 此算法正常工作的基础是图是0-indexed的。但是获得的结果Component数组和Belong数组是1-indexed */ int main() {    int n,m;     //n个点，m条边    while(scanf("%d %d",&n,&m)==2)     {         int a,b;         int ncount[M]={0}; //定义收缩之后连通分量的出度         for(int i=0;i<m;i++)         {             scanf("%d %d",&a,&b);             a--,b--;             Edge[a].push_back(b);         }         int  N=n;         solve(N);         reshape(N);         int time=0,ans=0;         for(int i=1;i<=ComponetNumber;i++)             if(!bcount[i])             {                 ans=Component[i].size();                 time++;             }         if(time==1)             printf("%d/n",ans);         else         printf("%d/n",0);     }     return 0; }

Kosaraju

#include "cstdlib" #include "cctype" #include "cstring" #include "cstdio" #include "cmath" #include "algorithm" #include "vector" #include "string" #include "iostream" #include "sstream" #include "set" #include "queue" #include "stack" #include "fstream" #include "strstream" using namespace std; #define M 20000 bool vis[M];                 //遍历数组 int post[M];                 //时间戳对应的点 int timestamp;               //时间戳 int ComponetNumber=0;        //有向图强连通分量个数 int Belong[M]; vector Edge[M];        //邻接表表示 vector Opp[M];         //原图的反图 vector Component[M];   //获得强连通分量结果 bool bcount[M]={0};void dfs(int u) {             //第一个dfs确定时间戳     vis[u] = true;     for(int i=0;i<Edge[u].size();i++) {         if(vis[ Edge[u][i]])    continue;         //cout<<Edge[u][i]<<endl;         dfs(Edge[u][i]);     }     //cout<<"timestamp    "<<timestamp<<"       "<<u<<endl;        post[timestamp++] = u; }void dfs2(int u) {      //第二个反边dfs确定连通块     vis[u] = true;     Component[ComponetNumber].push_back(u);     Belong[u]=ComponetNumber;     for(int i=0;i<Opp[u].size();i++)     {         int v = Opp[u][i];         if(vis[v])  continue;         dfs2(v);     } }void Kosaraju(int n) {     memset(Belong,0,sizeof(Belong));     memset(vis,0,sizeof(vis));     timestamp = 0;     for(int i=0;i<n;i++) {         if(vis[i])    continue;         dfs(i);     }     memset(vis,0,sizeof(vis));     ComponetNumber++;     for(int i=n-1;i>=0;i--) {//按时间戳从大到小搜         if(vis[post[i]])    continue;         Component[ComponetNumber].clear();         dfs2(post[i]);         ComponetNumber++;     }     ComponetNumber--;      //最后我们把块加了1。。所以要减掉 } void reshape(int N)     //缩点形成新图,N为图中的点数 {     //bool ComDIG[M][M];     //memset(ComDIG,0,sizeof(ComDIG));     for( int i =0 ; i<  N;i++)       //一个顶点是i     {         for(int j=0;j<Edge[i].size();j++) //另外一个顶点是edge[i][j]         {             if(Belong[i]!=Belong[Edge[i][j]])                 bcount[Belong[i]]=true;             //    ComDIG[Belong[i]][Belong[Edge[i][j]]]=true;         }     }     /*for( int i =1 ;i<=ComponetNumber;i++)     {         for(int j=1;j<=ComponetNumber;j++)         {             if(ComDIG[i][j])                 cout<<i<<"     "<<j<<endl;         }     }*/}// 此算法工作的前提仍然是0-indexed int main() {    int n,m;     //n个点，m条边    while(scanf("%d %d",&n,&m)==2)     {         int a,b;         int ncount[M]={0}; //定义收缩之后连通分量的出度         for(int i=0;i<m;i++)         {             scanf("%d %d",&a,&b);             a--,b--;             Edge[a].push_back(b);             Opp[b].push_back(a);         }         int  N=n;         Kosaraju(N);         reshape(N);         int time=0,ans=0;         for(int i=1;i<=ComponetNumber;i++)             if(!bcount[i])             {                 ans=Component[i].size();                 time++;             }             if(time==1)                 printf("%d/n",ans);             else                 printf("%d/n",0);     }     return 0; }

Tarjan

#include "cstdlib" #include "cctype" #include "cstring" #include "cstdio" #include "cmath" #include "algorithm" #include "vector" #include "string" #include "iostream" #include "sstream" #include "set" #include "queue" #include "stack" #include "fstream" #include "strstream" using namespace std;#define  M 10005             //题目中可能的最大点数       int STACK[M],top=0;          //Tarjan 算法中的栈 bool InStack[M];             //检查是否在栈中 int DFN[M];                  //深度优先搜索访问次序 int Low[M];                  //能追溯到的最早的次序 int ComponetNumber=0;        //有向图强连通分量个数 int Belong[M]; int Index=0;                 //索引号 vector Edge[M];        //邻接表表示 vector Component[M];   //获得强连通分量结果 //bool ComDIG[M][M]; bool bcount[M]={0};void Tarjan(int i) {     int j;     DFN[i]=Low[i]=Index++;     InStack[i]=true;     STACK[++top]=i;     for (int e=0;e<Edge[i].size();e++)     {         j=Edge[i][e];         if (DFN[j]==-1)         {             Tarjan(j);             Low[i]=min(Low[i],Low[j]);         }         else if (InStack[j])             Low[i]=min(Low[i],DFN[j]);     }     if (DFN[i]==Low[i])     {         //cout<<"TT    "<<i<<"   "<<Low[i]<<endl;         ComponetNumber++;         do         {             j=STACK[top--];             InStack[j]=false;             Component[ComponetNumber].push_back(j);             Belong[j]=ComponetNumber;         }         while (j!=i);     } }void solve(int N)     //此图中点的个数，注意是0-indexed！ {     memset(STACK,-1,sizeof(STACK));     memset(InStack,0,sizeof(InStack));     memset(DFN,-1,sizeof(DFN));     memset(Low,-1,sizeof(Low));     memset(Belong,0,sizeof(Belong));     for(int i=0;i<N;i++)         if(DFN[i]==-1)             Tarjan(i);    } void reshape(int N)     //缩点形成新图,N为图中的点数 {     //bool ComDIG[M][M];     //memset(ComDIG,0,sizeof(ComDIG));     for( int i =0 ; i<  N;i++)       //一个顶点是i     {         for(int j=0;j<Edge[i].size();j++) //另外一个顶点是edge[i][j]         {             if(Belong[i]!=Belong[Edge[i][j]])                 bcount[Belong[i]]=true;             //    ComDIG[Belong[i]][Belong[Edge[i][j]]]=true;         }     }     /*for( int i =1 ;i<=ComponetNumber;i++)     {         for(int j=1;j<=ComponetNumber;j++)         {             if(ComDIG[i][j])                 cout<<i<<"     "<<j<<endl;         }     }*/} /* 此算法正常工作的基础是图是0-indexed的。 */ int main() {    int n,m;     //n个点，m条边    while(scanf("%d %d",&n,&m)==2)     {         int a,b;         int ncount[M]={0}; //定义收缩之后连通分量的出度         for(int i=0;i<m;i++)         {             scanf("%d %d",&a,&b);             a--,b--;             Edge[a].push_back(b);         }         int  N=n;         solve(N);         reshape(N);     /*    for(int i=1;i<=ComponetNumber;i++)         {             for(int j=0;j<ComponetNumber;j++)             {                 if(ComDIG[i][j])                     bcount[i]++;             }         }*/         int time=0,ans=0;         for(int i=1;i<=ComponetNumber;i++)             if(!bcount[i])             {                 ans=Component[i].size();                 time++;             }         if(time==1)             printf("%d/n",ans);         else         printf("%d/n",0);     }     return 0; }