这道题可以把边都反着存一遍，从终点开始深搜，然后把到不了的点 和它们所指向的点都去掉。
最后在剩余的点里跑一遍spfa就可以了。
——代码
#include <cstdio>
#include <cstring>
#include <queue> const int maxn = ;
int n, m, s, t, cnt1, cnt2;
int dis[maxn];
int next1[ * maxn], to1[ * maxn], head1[ * maxn],
next2[ * maxn], to2[ * maxn], head2[ * maxn];
bool vis[maxn], b[maxn], f[maxn], flag; inline void add1(int x, int y)
{
to1[cnt1] = y;
next1[cnt1] = head1[x];
head1[x] = cnt1++;
} inline void add2(int x, int y)
{
to2[cnt2] = y;
next2[cnt2] = head2[x];
head2[x] = cnt2++;
} inline void dfs(int u)
{
int i, v;
if(u == s) flag = ;
vis[u] = ;
for(i = head2[u]; i != -; i = next2[i])
{
v = to2[i];
if(!vis[v]) dfs(v);
}
} inline void spfa(int u)
{
int i, j;
std::queue <int> q;
q.push(u);
f[u] = ;
dis[u] = ;
while(!q.empty())
{
i = q.front();
q.pop();
f[i] = ;
for(j = head1[i]; j != -; j = next1[j])
if(!b[i] && !b[to1[j]] && dis[to1[j]] > dis[i] + )
{
dis[to1[j]] = dis[i] + ;
if(!f[to1[j]])
{
f[to1[j]] = ;
q.push(to1[j]);
}
}
}
} int main()
{
int i, j, x, y;
memset(head1, -, sizeof(head1));
memset(head2, -, sizeof(head2));
memset(dis, / , sizeof(dis));
scanf("%d %d", &n, &m);
for(i = ; i <= m; i++)
{
scanf("%d %d", &x, &y);
add1(x, y);
add2(y, x);
}
scanf("%d %d", &s, &t);
dfs(t);
if(!flag)
{
printf("-1");
return ;
}
for(i = ; i <= n; i++)
if(!vis[i])
{
b[i] = ;
for(j = head2[i]; j != -; j = next2[j]) b[to2[j]] = ;
}
spfa(s);
printf("%d", dis[t]);
return ;
}