Codeforces Round #490 (Div. 3) E. Reachability from the Capital (tarjan缩点)

• 题意:有\(n\)个点,\(m\)条边,问你最少加多少条边,使得点\(s\)可以走到任何一个点.

• 题解:我们直接用tarjan缩点后,判断除了\(s\)以外强连通分量的入度为\(0\)的个数即可.

• 代码:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
  ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
   
  int n,m,s;
  int dfn[N],low[N],timestamp;
  int stk[N],top;
  bool in_stk[N];
  int id[N],scc_cnt,sz[N];
  int din[N];
  
  vector<int> edge[N];
  
  void tarjan(int u){
      dfn[u]=low[u]=++timestamp;
      stk[++top]=u,in_stk[u]=true;
      for(auto w:edge[u]){
          if(!dfn[w]){
              tarjan(w);
              low[u]=min(low[u],low[w]);
          }
          else if(in_stk[w]) low[u]=min(low[u],dfn[w]);
      }
  
      if(dfn[u]==low[u]){
          ++scc_cnt;
          int y;
          do{
              y=stk[top--];
              in_stk[y]=false;
              id[y]=scc_cnt;
              sz[scc_cnt]++;
          }while(y!=u);
      }
  
  }
   
  int main(){
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      cin>>n>>m>>s;
      rep(i,1,m){
          int u,v;
          cin>>u>>v;
          edge[u].pb(v);
      }
  
      rep(i,1,n){
          if(!dfn[i]) tarjan(i);
      }
   
      rep(i,1,n){
          for(auto w:edge[i]){
              int u=id[i];
              int v=id[w];
              if(u!=v){
                  din[v]++;
              }
          }
      }
  
      int ori=id[s];
      int ans=0;
      rep(i,1,scc_cnt){
          if(i==ori) continue;
          if(din[i]==0) ans++;
      }
  
      cout<<ans<<'\n';
   
      return 0;
  }