結果

問題 No.2115 Making Forest Easy
ユーザー 👑 ygussany
提出日時 2022-10-29 00:03:29
言語 C
(gcc 13.3.0)
結果
TLE  
実行時間 -
コード長 2,548 bytes
コンパイル時間 303 ms
コンパイル使用メモリ 33,864 KB
実行使用メモリ 86,592 KB
最終ジャッジ日時 2024-07-06 03:20:20
合計ジャッジ時間 3,949 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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ファイルパターン 結果
sample AC * 2
other AC * 1 TLE * 1 -- * 48
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <stdio.h>

const int Mod = 998244353;

typedef struct Edge {
	struct Edge *next;
	int v, id;
	long long dp[1001];
} edge;

int main()
{
	int i, N, M, u, w, A[5001];
	edge *adj[5001] = {}, e[10001], *p;
	scanf("%d", &N);
	for (u = 1; u <= N; u++) scanf("%d", &(A[u]));
	for (i = 0; i < N - 1; i++) {
		scanf("%d %d", &u, &w);
		e[i*2].v = w;
		e[i*2+1].v = u;
		e[i*2].id = i * 2;
		e[i*2+1].id = i * 2 + 1;
		e[i*2].next = adj[u];
		e[i*2+1].next = adj[w];
		adj[u] = &(e[i*2]);
		adj[w] = &(e[i*2+1]);
	}
	
	int par[5001] = {}, pred[5001], q[5001], head, tail;
	q[0] = 1;
	par[1] = 1;
	pred[1] = -1;
	for (head = 0, tail = 1; head < tail; head++) {
		u = q[head];
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			if (par[w] == 0) {
				par[w] = u;
				pred[w] = p->id;
				q[tail++] = w;
			}
		}
	}
	
	int k;
	long long sum;
	for (head--; head >= 1; head--) {
		u = q[head];
		i = pred[u];
		
		if (adj[u]->next == NULL) {
			for (k = 0; k < A[u]; k++) e[i].dp[k] = 1;
			for (k = A[u]; k <= 1000; k++) e[i].dp[k] = 2;
			continue;
		}
		
		for (p = adj[u], sum = 1; p != NULL; p = p->next) {
			if (p->v == par[u]) continue;
			sum = sum * p->dp[1000] % Mod;
		}
		e[i].dp[0] = sum;
		for (k = 1; k < A[u]; k++) e[i].dp[k] = e[i].dp[0];
		for (k = A[u]; k <= 1000; k++) {
			e[i].dp[k] = e[i].dp[0];
			for (p = adj[u], sum = 1; p != NULL; p = p->next) {
				if (p->v == par[u]) continue;
				sum = sum * p->dp[k] % Mod;
			}
			e[i].dp[k] = (e[i].dp[0] + sum) % Mod;
		}
	}
	for (head = 1; head < tail; head++) {
		w = q[head];
		i = pred[w] ^ 1;
		u = par[w];
		
		if (adj[u]->next == NULL) {
			for (k = 0; k < A[u]; k++) e[i].dp[k] = 1;
			for (k = A[u]; k <= 1000; k++) e[i].dp[k] = 2;
			continue;
		}
		
		for (p = adj[u], sum = 1; p != NULL; p = p->next) {
			if (p->id == (i ^ 1)) continue;
			sum = sum * p->dp[1000] % Mod;
		}
		e[i].dp[0] = sum;
		for (k = 1; k < A[u]; k++) e[i].dp[k] = e[i].dp[0];
		for (k = A[u]; k <= 1000; k++) {
			e[i].dp[k] = e[i].dp[0];
			for (p = adj[u], sum = 1; p != NULL; p = p->next) {
				if (p->id == (i ^ 1)) continue;
				sum = sum * p->dp[k] % Mod;
			}
			e[i].dp[k] = (e[i].dp[0] + sum) % Mod;
		}
	}
	
	long long ans = 0, total[1001];
	for (u = 1; u <= N; u++) {
		for (k = 0; k <= 1000; k++) {
			for (p = adj[u], total[k] = 1; p != NULL; p = p->next) total[k] = total[k] * p->dp[k] % Mod;
			if (k > 0) ans += (total[k] - total[k-1] + Mod) * ((k >= A[u])? k: A[u]) % Mod;
		}
		ans += total[0] * A[u] % Mod;
	}
	printf("%lld\n", ans % Mod);
	fflush(stdout);
	return 0;
}
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