CodeForces - 999E

题意：找有多少个单向联通的块
题解：由s开始dfs找到无法从s到达的点，虚拟的连上一条边，然后暴力dfs所有从s无法到达的点，当这个点能个够到达一个存在虚拟边的点时，把虚拟边删除。最后看一下有多少个虚拟边就是答案。
AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5003;
int edgesNum = 0, Head[maxn];
bool vis[maxn], needLink[maxn];
struct EDGE {
    int to, next;
} edges[maxn];

void addEdge(int u, int v) {
    edges[edgesNum].to = v;
    edges[edgesNum].next = Head[u];
    Head[u] = edgesNum++;
}
void dfs(int u) {
    vis[u] = 1;
    for(int e = Head[u]; e != - 1; e = edges[e].next) {
        int v = edges[e].to;
        if(!vis[v]) {
            dfs(v);
        }
    }
}
void dfs1(int u) {
    vis[u] = 1;
    for(int e = Head[u]; e != -1; e = edges[e].next) {
        int v = edges[e].to;
        if(!vis[v]) {
            needLink[v] = false;
            dfs1(v);
        }
    }
}

int main() {
    memset(vis, false, sizeof(vis));
    memset(needLink, false, sizeof(needLink));
    memset(Head, -1, sizeof(Head));
    int n, m, s, u, v;
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &u, &v);
        addEdge(u, v);
    }
    dfs(s);
    for(int i = 1; i <= n; i++) {
        if(!vis[i]) {
            needLink[i] = true;
        }
    }
    memset(vis, false, sizeof(vis));
    for(int i = 1; i <= n; i++) {
        if(needLink[i]) {
            dfs1(i);
            memset(vis, false, sizeof(vis));
        }
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        if(needLink[i]) ans++;
    }
    printf("%d\n", ans);
    return 0;
}