【题解】LuoGu2296：寻找道路

原题传送门
细节挺多的，如果心情浮躁的话就很难调出来
从终点跑反向边跑出能达到的点，这样就知道了哪些点无法到达终点
用这些点跑反向边跑出出点无法和终点联通的点，这些点也是不能到达的
剩下的点就是可以走的点，用$bfs$就可以了

Code：

#include <bits/stdc++.h>
#define maxn 200010
using namespace std;
struct Edge{
	int to, next;
}edge1[maxn << 1], edge2[maxn << 1];
int num1, num2, head1[maxn], head2[maxn], vis[maxn], flag[maxn], dis[maxn], n, m, s, t;
queue <int> q;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge1(int x, int y){ edge1[++num1] = (Edge){y, head1[x]}, head1[x] = num1; }
void addedge2(int x, int y){ edge2[++num2] = (Edge){y, head2[x]}, head2[x] = num2; }

void dfs(int u){
	vis[u] = 1;
	for (int i = head2[u]; i; i = edge2[i].next){
		int v = edge2[i].to;
		if (!vis[v]) dfs(v);
	}
}

int main(){
	n = read(), m = read();
	for (int i = 1; i <= m; ++i){
		int x = read(), y = read();
		addedge1(x, y), addedge2(y, x);
	}
	s = read(), t = read();
	dfs(t);
	for (int i = 1; i <= n; ++i)
		if (vis[i]){
			for (int j = head1[i]; j; j = edge1[j].next){
				int v = edge1[j].to;
				if (!vis[v]) flag[i] = 1;
			}
		}
	for (int i = 1; i <= n; ++i) if (flag[i]) vis[i] = 0;
	if (vis[s]) q.push(s);
	while (!q.empty()){
		int u = q.front();
		q.pop();
		for (int i = head1[u]; i; i = edge1[i].next){
			int v = edge1[i].to;
			if (vis[v] && !dis[v]){
				dis[v] = dis[u] + 1;
				q.push(v);
			}
		}
	}
	if (!dis[t]) puts("-1"); else printf("%d\n", dis[t]);
	return 0;
}