题意：求强联通分量
Tarjan算法
1 #include<iostream> 2 #include<stdio.h> 3 #include<stdlib.h> 4 #include<memory.h> 5 #include<string.h> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #define clc(a,b) memset(a,b,sizeof(a)) 10 typedef long double ld; 11 typedef long long ll; 12 const int N = 30; 13 const double eps=1e-9; 14 const int inf=-100000; 15 const int maxn=1e5+10; 16 const int num=10001; 17 const double Pi=acos(-1); 18 using namespace std; 19 20 struct gragh 21 { 22 int to; 23 int next; 24 } V[num]; 25 26 bool instack[num]= {false}; 27 int low[num]= {0},DFN[num]= {0},Stap[num]= {0},Belong[num]= {0}; 28 int answer=0; 29 int Dindex,stop,Bcnt; 30 int head[num]; 31 int edge; 32 33 void add(int a,int b) 34 { 35 V[edge].to=b; 36 V[edge].next=head[a]; 37 head[a]=edge++; 38 39 } 40 void tarjan(int u) 41 { 42 int v; 43 DFN[u]=low[u]=++Dindex; 44 instack[u]=true; 45 Stap[stop++]=u; 46 for(int i=head[u]; i!=-1; i=V[i].next) 47 { 48 v=V[i].to; 49 if(!DFN[v]) 50 { 51 tarjan(v); 52 if(low[v]<low[u]) 53 low[u]=low[v]; 54 } 55 else if(instack[v]&&DFN[v]<low[u]) 56 low[u]=DFN[v]; 57 } 58 if(DFN[u]==low[u]) 59 { 60 int sum=0; 61 Bcnt++; 62 do 63 { 64 v=Stap[--stop]; 65 instack[v]=false; 66 Belong[v]=Bcnt; 67 sum++; 68 } 69 while(v!=u); 70 cout<<sum<<endl; 71 if(sum!=0) 72 answer+=(sum*(sum-1))/2; 73 } 74 } 75 void solve(int N) 76 { 77 stop=Bcnt=Dindex=0; 78 // clc(DFN,0); 79 for(int i=1; i<=N; i++) 80 if(!DFN[i]) 81 tarjan(i); 82 } 83 84 int main() 85 { 86 int n,m; 87 int a,b; 88 cin>>n>>m; 89 clc(head,-1); 90 edge=0; 91 for(int i=0; i<m; i++) 92 { 93 cin>>a>>b; 94 add(a,b); 95 } 96 solve(n); 97 cout<<answer<<endl; 98 return 0; 99 }