AtCoder Beginner Contest 132
E
Hopscotch Addict
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原题
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翻译中…

Problem Statement

翻译中…

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph $G$. $G$ consists of $N$ vertices numbered $1$ to $N$, and $M$ edges. The $i$-th edge points from Vertex $u_i$ to Vertex $v_i$.

翻译中…

First, Ken stands on Vertex $S$. He wants to reach Vertex $T$ by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

翻译中…

Determine if he can reach Vertex $T$ by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex $T$. Note that visiting Vertex $T$ in the middle of a ken-ken-pa does not count as reaching Vertex $T$ by repeating ken-ken-pa.

翻译中…

Constraints

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  • $2 \leq N \leq 10^5$
  • $0 \leq M \leq \min(10^5, N (N-1))$
  • $1 \leq u_i, v_i \leq N(1 \leq i \leq M)$
  • $u_i \neq v_i (1 \leq i \leq M)$
  • If $i \neq j$, $(u_i, v_i) \neq (u_j, v_j)$.
  • $1 \leq S, T \leq N$
  • $S \neq T$
翻译中…

Input

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Input is given from Standard Input in the following format:

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N M
u_1 v_1
:
u_M v_M
S T

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Output

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If Ken cannot reach Vertex $T$ from Vertex $S$ by repeating ken-ken-pa, print $-1$. If he can, print the minimum number of ken-ken-pa needed to reach vertex $T$.

翻译中…

样例

Input #1
4 4
1 2
2 3
3 4
4 1
1 3
Output #1
2
Ken can reach Vertex $3$ from Vertex $1$ in two ken-ken-pa, as follows: $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ in the first ken-ken-pa, then $4 \rightarrow 1 \rightarrow 2 \rightarrow 3$ in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.
Input #2
3 3
1 2
2 3
3 1
1 2
Output #2
-1
Any number of ken-ken-pa will bring Ken back to Vertex $1$, so he cannot reach Vertex $2$, though he can pass through it in the middle of a ken-ken-pa.
Input #3
2 0
1 2
Output #3
-1
Vertex $S$ and Vertex $T$ may be disconnected.
Input #4
6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6
Output #4
2
题目信息
题号
abc132_e
时间限制
2 s
内存限制
1024 MB

题面与样例版权归 AtCoder 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。