Gifts by the List CodeForces

大意: 给定森林, 要求构造一个表, 满足对于每个$x$, 表中第一次出现的$x$的祖先(包括$x$)是$a_x$.

刚开始还想着直接暴力分块优化一下连边, 最后按拓扑序输出...

实际上可以发现$a_x$要么等于$x$, 要么等于$fa_x$, 直接dfs一边就好了...

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif


int n, m;
vector<int> g[N];
int s[N], vis[N], a[N];
void dfs(int x) {
	if (vis[x]) return;
	vis[x] = 1;
	for (int y:g[x]) {
		if (a[y]!=y&&a[y]!=a[x]) puts("-1"),exit(0);
		dfs(y);
	}
	if (a[x]==x) s[++*s]=x;
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,m) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v);
	}
	REP(i,1,n) scanf("%d", a+i);
	REP(i,1,n) dfs(i);
	printf("%d
", *s);
	REP(i,1,*s) printf("%d
", s[i]);
}