solution-2022 CCPC Guilin J. Permutation Puzzle

題解：2022 CCPC 桂林站 J 題題解

模擬賽 T3 放了這道題人均場切了。我沒刪除錯爆零了。

首先按所有限制連邊 \(u_i \to v_i\)。題目保證了這是一張有向無環圖。

我們肯定是隻能按照某種拓撲序來填。

有一個非常顯然的策略是在拓撲排序中按照每個點的後繼節點的最小值為第一關鍵字，更小的先填。

這非常符合直覺，也確實是對的。因為無論如何你都要在當前最小值之前到那。

#include<bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const int mininf = 1e9 + 7;
#define int long long
#define pb emplace_back
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void write(int x){if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0');}
#define put() putchar(' ')
#define endl puts("")
const int MAX = 2e5 + 10;

vector <int> g[MAX];
vector <int> g2[MAX];
bool vis[MAX];
int a[MAX];

void dfs(int u){
	vis[u] = 1;
	for(int v : g[u]){
		if(!vis[v])	dfs(v);
		a[u] = min(a[u], a[v]);
	}
}

int deg[MAX];
int deg2[MAX];
int b[MAX];
int to[MAX];
bool fl[MAX];

void solve(){
	int n = read(), m = read();
	for(int i = 1; i <= n; i++)a[i] = b[i] = deg[i] = deg2[i] = vis[i] = to[i] = fl[i] = 0;
	for(int i = 1; i <= n; i++)g[i].clear(),g2[i].clear();
	for(int i = 1; i <= n; i++){
		a[i] = read();
		if(a[i] == 0)	a[i] = inf;
		else b[i] = a[i], to[a[i]] = i;
	}
	for(int i = 1; i <= m; i++){
		int u = read(), v = read();
		g[u].pb(v);
		g2[v].pb(u);
		deg[v]++;
		deg2[u]++;
	}
	queue <int> que2;
	for(int i = 1; i <= n; i++){
		if(!deg2[i])	que2.push(i);
	}
	while(!que2.empty()){
		int u = que2.front();
		que2.pop();
		for(int v : g2[u]){
			deg2[v]--;
			a[v] = min(a[v], a[u]);
			if(!deg2[v])	que2.push(v);
		}
	}
	// for(int i = 1; i <= n; i++){
		// write(a[i]), put();
	// }endl;
	
	priority_queue <pair <int, int> > que;
	
	// write(deg[13]), endl;
	// for(int v : g[13]){
		// write(a[v]), endl;
	// }
	
	for(int i = 1; i <= n; i++){
		// if(b[i] == 700){
			// write(a[13]), endl;
			// for(int v : g2[i])	write(v), put();
			// endl;
		// }
		if(!deg[i]){
			
			if(b[i] == a[i]){
				// if(b[i] == 700)	puts("fjk");
				// write(g[i] == 700)	puts("kf");
				fl[b[i]] = 1;
			}else que.push(make_pair(-a[i], i));
		}
	}
	// write(que.top().first), endl;
	for(int i = 1; i <= n; i++){
		if(to[i]){
			if(!fl[i]){
				// write(i), endl;
				puts("-1");
				
				for(int i = 1; i <= n; i++)a[i] = b[i] = deg[i] = deg2[i] = vis[i] = to[i] = fl[i] = 0;
				for(int i = 1; i <= n; i++)g[i].clear(),g2[i].clear();
				return ;
			}	
			if(!que.empty() and -que.top().first <= i){
				puts("-1");
				
				for(int i = 1; i <= n; i++)a[i] = b[i] = deg[i] = deg2[i] = vis[i] = to[i] = fl[i] = 0;
				for(int i = 1; i <= n; i++)g[i].clear(),g2[i].clear();
				// write(i), put();
				return ;
			}	
			int u = to[i];
			b[u] = i;
			for(int v : g[u]){
				deg[v]--;
				if(!deg[v]){
					if(a[v] == b[v]){
						fl[b[v]] = 1;
					}else{
						que.push(make_pair(-a[v], v));
					}
				}
			}
		}else{
			if(que.empty())	{
				
				puts("-1");
				for(int i = 1; i <= n; i++)a[i] = b[i] = deg[i] = deg2[i] = vis[i] = to[i] = fl[i] = 0;
				for(int i = 1; i <= n; i++)g[i].clear(),g2[i].clear();
				return ;
			}
			int u = que.top().second;
			b[u] = i;
			que.pop();
			for(int v : g[u]){
				deg[v]--;
				if(!deg[v]){
					if(a[v] == b[v]){
						fl[b[v]] = 1;
					}else{
						que.push(make_pair(-a[v], v));
					}
				}
			}
		}
	}
	for(int i = 1; i <= n; i++){
		write(b[i]), put();
	}
	endl;
}

signed main(){
	// freopen("permutation.in", "r", stdin);
	// freopen("permutation.out", "w", stdout);
	int t = read();
	while(t--)	solve();
	return 0;
}