E. Reachability from the Capital[强联通]

E. Reachability from the Capital

题意:一张有向图,现在要求从S出发能到所有点,问至少加几条边

思路:强联通缩点后,判断入度为0的块有几个,S除外

#include<bits/stdc++.h>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
const int N=5005;
const int MOD=1e9+7;
vector <int> edge[N];
int n,m,SSS,cur,cnt,top,mp[N],DFN[N],LOW[N],s[N];
struct node{int a;int b;}p[N];
int in[N];
void tarjan(int u){
    DFN[u]=LOW[u]=++cur;
    s[++top]=u;
    for(int i=0;i<(int)edge[u].size();i++){
        int v=edge[u][i];
        if(!DFN[v]){
            tarjan(v);
            LOW[u]=min(LOW[u],LOW[v]);
        }
        else if(DFN[v] && !mp[v])   LOW[u]=min(LOW[u],DFN[v]);
    }
    if(DFN[u]==LOW[u]){
        int v=-1;
        cnt++;
        while(u!=v){
            v=s[top--];
            mp[v]=cnt;
//            cout << v<<" ";
        }
//        puts("************");
    }
}
int main(void){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin >>n>>m>>SSS;
    for(int i=1;i<=m;i++){
        int u,v;
        cin >>u>>v;
        p[i].a=u,p[i].b=v;
        edge[v].pb(u);
    }
    ll ans=0;

    for(int i=1;i<=n;i++)
        if(!DFN[i]) tarjan(i);
//    cout <<"cnt="<<cnt<<endl;
    for(int i=1;i<=m;i++){
        int a=p[i].a,b=p[i].b;
        if(mp[a]==mp[b])    continue;
        in[mp[b]]++;
    }

    for(int i=1;i<=cnt;i++){
        if(i==mp[SSS])    continue;
//        cout <<i << endl;
        if(in[i]==0)    ans++;
    }
    cout << ans << "\n";


    return 0;
}