CodeForces

嘛,首先把图缩点一下之后就很好搞了

考虑一下,什么情况下可以使得加边最少,并且s可以把所有点遍历到?

很显然,对于一个入度不为0的点,一定有前继节点....

于是乎,我们只要找到不满足有前继节点的节点就好了

详细的过程看代码...

#include<bits/stdc++.h>
#define MAXN 5005
using namespace std;

int n,m,tot,tot2,dx,tx;
int s,p,ed;
int sum;
int in[MAXN];
int h[MAXN],h2[MAXN],low[MAXN],dfn[MAXN],belong[MAXN];
bool vis[MAXN];
stack<int>q;

struct node{
    int from,to,next;
}e[MAXN<<1],e2[MAXN<<1];

void init(){
    tot = tot2 = dx = tx = sum = 0;
    memset(h,-1,sizeof(h));
    memset(h2,-1,sizeof(h2));
    memset(dfn,0,sizeof(dfn));
    memset(in,0,sizeof(in));
    memset(low,0,sizeof(low));
    memset(vis,false,sizeof(vis));
}

void add(int x,int y){
    tot++;
    e[tot].from = x;
    e[tot].to = y;
    e[tot].next = h[x];
    h[x] = tot;
}

void add2(int x,int y){
    tot2++;
    e2[tot2].from = x;
    e2[tot2].to = y;
    e2[tot2].next = h2[x];
    h2[x] = tot2;
}

int tarjan(int now){
    dx++;
    low[now] = dfn[now] = dx;
    q.push(now),vis[now] = true;
    for(int i = h[now];i != (-1);i = e[i].next){
        if(!dfn[e[i].to]){
            tarjan(e[i].to);
            low[now] = min(low[now] , low[e[i].to]);
        }
        else if(vis[e[i].to]){
            low[now] = min(low[now] , dfn[e[i].to]);
        }
    }
    if(dfn[now]==low[now]){
        int u;
        tx++;
        do{
            u = q.top();
            belong[u] = tx;
            q.pop();
            vis[u] = false;
        }while(u!=now);
    }
}

void solve(){
    for(int i = 1;i<=tx;i++){
        if(!in[i]&&i != belong[s]){
            sum++;
            in[i]++;
        }
    }
    cout<<sum<<endl;
}

int main(){
    init();
    cin>>n>>m>>s;
    for(int i = 1;i<=m;i++){
        int x,y;cin>>x>>y;
        add(x,y);
    }
    for(int i = 1;i<=n;i++)
        if(!dfn[i])tarjan(i);
    for(int i = 1;i<=m;i++){
        if(belong[e[i].from] == belong[e[i].to])continue;
        add2(belong[e[i].from] , belong[e[i].to]);
        in[belong[e[i].to]]++;
    }
    solve();
}