CSP2020 | T4

T3

假设有函数1，后有函数2为 ( imes v)，则函数1相当于做了 (v) 次。
利用这个性质可以把所有的操作变成加法操作。
考虑执行函数序列 (q_1,q_2,cdots,q_Q) ，设 (f_i) 为 (i) 函数做的乘法操作的值的积，则函数 (q_{Q-1}) 相当于做了 (f_{q_Q}) 次，函数 (q_i) 的实际执行次数 (cnt_{q_i} = prodlimits_{j=i+1}^{Q} f_{q_j}) 次。
可以先 DFS 求出 (f_i) ，再从 (Q) 到 (1) 扫一遍求出 (cnt_{q_i}) ，然后跑一遍拓扑排序，用 (cnt_u) 更新 (cnt_v) 即可。

#include <cstdio>
#include <cctype>
#include <cstring>
#define MAX_N (100000 + 5)
#define MAX_M (100000 + 5)
#define MAX_Q (100000 + 5)
#define MAX_SUM (1000000 + 5)

typedef long long LL;
const int mod = 998244353;
int n, m, Q;
int a[MAX_N];
struct Opt {
	int t, p, v;
	int c;
} o[MAX_N];
int h[MAX_N], tot;
struct Edge {
	int to, next;
} e[MAX_SUM];
int q[MAX_Q];
int in[MAX_M];
int f[MAX_M];
bool vis[MAX_M];
int cnt[MAX_M];
int st[MAX_M], l, r;

int Read() {
	int res = 0;
	char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) res = res * 10 + ch - '0', ch = getchar();
	return res;
}

void Write(int x) {
	if (x >= 10) Write(x / 10);
	putchar(x % 10 + '0');
}

void Add_Edge(int u, int v) {
	e[++tot].to = v;
	e[tot].next = h[u];
	h[u] = tot;
	++in[v];
}

void DFS(int u) {
	if (vis[u]) return;
	vis[u] = 1;
	int v;
	for (int i = h[u]; i; i = e[i].next) {
		v = e[i].to;
		DFS(v);
		f[u] = (LL)f[u] * f[v] % mod;
	}
}

int main() {
	n = Read();
	for (int i = 1; i <= n; ++i)
		a[i] = Read();
	m = Read();
	for (int i = 1; i <= m; ++i) {
		o[i].t = Read();
		switch (o[i].t) {
		case 1:
			o[i].p = Read(), o[i].v = Read();
			break;
		case 2:
			o[i].v = Read();
			break;
		default:
			o[i].c = Read();
			for (int j = 1; j <= o[i].c; ++j)
				Add_Edge(i, Read());
		}
	}
	Q = Read();
	for (int i = 1; i <= Q; ++i)
		q[i] = Read();
	for (int i = 1; i <= m; ++i)
		f[i] = o[i].t == 2 ? o[i].v : 1;
	for (int i = 1; i <= m; ++i)
		if (!in[i]) DFS(i);
	int prod = 1;
	for (int i = Q; i; --i) {
		cnt[q[i]] = (cnt[q[i]] + prod) % mod;
		prod = (LL)prod * f[q[i]] % mod;
	}
	l = 1;
	for (int i = 1; i <= m; ++i)
		if (!in[i]) st[++r] = i;
	int u, v, tmp;
	while (l <= r) {
		u = st[l++];
		tmp = cnt[u];
		for (int i = h[u]; i; i = e[i].next) {
			v = e[i].to;
			cnt[v] = (cnt[v] + tmp) % mod;
			tmp = (LL)tmp * f[v] % mod;
			--in[v];
			if (!in[v]) st[++r] = v;
		}
	}
	for (int i = 1; i <= n; ++i)
		a[i] = (LL)a[i] * prod % mod;
	for (int i = 1; i <= m; ++i)
		if (o[i].t == 1) 
			a[o[i].p] = (a[o[i].p] + (LL)cnt[i] * o[i].v) % mod;
	for (int i = 1; i <= n; ++i)
		Write(a[i]), putchar(' ');
	return 0;
}