#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=1005, Q=N*100, oo=~0u>>2;

int q[Q], front, tail, d[N], vis[N], ihead[N], cnt, n, x;

struct ED { int to, next; }e[Q];

void add(int u, int v) {

    e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v;

}

int spfa() {

    q[tail++]=1;

    for1(i, 2, 1005) d[i]=oo;

    vis[1]=1;

    while(front!=tail) {

        int v, u=q[front++]; if(front==Q) front=0; vis[u]=0;

        for(int i=ihead[u]; i; i=e[i].next) if(d[v=e[i].to]>d[u]+1) {

            d[v]=d[u]+1;

            if(!vis[v]) {

                vis[v]=1;

                q[tail++]=v; if(tail==Q) tail=0;

            }

        }

    }

    return d[x];

}

int main() {

    read(n); read(x);

    for1(i, 1, n) {

        int u=getint(), v=getint();

        add(u, v);

    }

    int ans=spfa();

    if(ans==oo) puts("-1");

    else print(ans+1);

    return 0;

}

奶牛们至少要交换4次，先用1去交换3，再用3去交换2，最后用2交换得到5．

Silver