The 2025 Shanghai Collegiate Programming Contest - I. 真相（类似于树上背包的转移）

思路讲解

递推公式。

auto dfs=[&](auto&& self,ll u,ll p)->void{
	dp[u][0]=1;		// 每一个人说真话这种情况一定是存在的
	for(auto &v:g[u]){
		if(v==p) continue;
		self(self,v,u);
		vector<ll> tmp(N+5);
		ROF(i,sz[u],0){
			FOR(j,0,sz[v]){
				tmp[i+j]+=dp[u][i]*dp[v][j]%mod;
				tmp[i+j]%=mod;
			}
		}
		dp[u]=tmp;
		sz[u]+=sz[v];
	}
	// dp[u][A[u]-1] 就是如果u说真话，dp[u][A[u]]的值
	// 因为u说真话，那么情况已经定下来了，其余的就是dp[u][A[u]-1]
	dp[u][A[u]]=A[u]?dp[u][A[u]-1]:0;  // 对于任何节点来说，A[u]=0都是悖论
};

AC代码

https://codeforces.com/gym/105992/submission/328455295

源代码

// Problem: I. 真相
// Contest: Codeforces - The 2025 Shanghai Collegiate Programming Contest
// URL: https://codeforces.com/gym/105992/problem/I
// Memory Limit: 512 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(5e3)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e6+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod= 998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);

ll N,M,Q,X,K,T,lT,A[MAXN];
/*

*/
inline void init(){
	
}
inline void solve(){
	cin>>N;
	init();
	FOR(i,1,N){
		cin>>A[i];
	}
	vector g(N+5,vector<ll>());
	vector dp(N+5,vector<ll>(N+5));
	FOR(i,1,N-1){
		ll u,v;
		cin>>u>>v;
		g[u].pb(v);g[v].pb(u);
	}
	vector<ll> sz(N+5,1);
	auto dfs=[&](auto&& self,ll u,ll p)->void{
		dp[u][0]=1;		// 每一个人说真话这种情况一定是存在的
		for(auto &v:g[u]){
			if(v==p) continue;
			self(self,v,u);
			vector<ll> tmp(N+5);
			ROF(i,sz[u],0){
				FOR(j,0,sz[v]){
					tmp[i+j]+=dp[u][i]*dp[v][j]%mod;
					tmp[i+j]%=mod;
				}
			}
			dp[u]=tmp;
			sz[u]+=sz[v];
		}
		dp[u][A[u]]=A[u]?dp[u][A[u]-1]:0;
	};
	dfs(dfs,1,1);
	ll ans=0;
	FOR(i,0,N){
		ans+=dp[1][i];ans%=mod;
	}
// #ifdef LOCAL
    // FOR(i,1,N){
    	// FOR(j,0,N){
    		// cerr<<dp[i][j]<<" ";
    	// }
    	// cerr<<"\n";
    // }
// #endif
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--)
		solve();
	return 0;
}
/*
AC
https://codeforces.com/gym/105992/my
*/

心路历程（WA，TLE，MLE……）