原題傳送門
細節挺多的,如果心情浮躁的話就很難調出來
從終點跑反向邊跑出能達到的點,這樣就知道了哪些點無法到達終點
用這些點跑反向邊跑出出點無法和終點聯通的點,這些點也是不能到達的
剩下的點就是可以走的點,用就可以了
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;
}