【树上DFS】Tree and Polynomials

http://codeforces.com/gym/101372

D

push1[i][k]：所有操作1总共要让节点i下推多少系数k

push2[i][k]：所有操作2总共要让节点i上推多少系数k

sum1[i][k]：所有操作1节点i要计算多少系数k

sum2[i][k]：所有操作2节点i要计算多少系数k

遍历k从1~20跑dfs处理出所有sum1，sum2数据以及每个点的深度，最后统一计算多项式每一项

注意用memset会超时

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int n, k, q, e, rt;
const int maxn = (int)1e5 + 10;
typedef long long ll;
const ll mod = (ll)1e9 + 7;
struct Edge {
	int to, next;
} es[maxn];
int head[maxn];
void add(int u, int v) {
	es[e].to = v;
	es[e].next = head[u];
	head[u] = e++;
}
ll push1[maxn][30], push2[maxn][30], sum1[maxn][30], sum2[maxn][30];
int deep[maxn];
ll dfs(int u, int k, int d, ll tmp) {
	deep[u] = d;
	sum1[u][k] = (push1[u][k] + tmp) % mod;
	sum2[u][k] = push2[u][k];
	for (int i = head[u]; ~i; i = es[i].next) {
		int v = es[i].to;
		sum2[u][k] = (sum2[u][k] + dfs(v, k, d + 1, (tmp + push1[u][k]) % mod)) % mod;
	}
	return sum2[u][k];
}
ll q_pow(ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) {
			ans = ans * a % mod;
		}
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}
int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		e = 0;
		scanf("%d%d", &n, &k);
		for (int i = 0; i <= n; i++) {
			head[i] = -1;
		}
		for (int i = 1, fa; i <= n; i++) {
			scanf("%d", &fa);
			if (!fa)
				rt = i;
			else
				add(fa, i);
			for (int j = 0; j <= k; j++) {
				push1[i][j] = push2[i][j] = 0;
			}
		}
		scanf("%d", &q);
		while (q--) {
			int op, v;
			scanf("%d%d", &op, &v);
			if (op == 1) {
				for (int i = 0; i <= k; i++) {
					ll qs;
					scanf("%lld", &qs);
					push1[v][i] = (push1[v][i] + qs) % mod;
				}
			}
			else {
				for (int i = 0; i <= k; i++) {
					ll qs;
					scanf("%lld", &qs);
					push2[v][i] = (push2[v][i] + qs) % mod;
				}
			}
		}
		for (int i = 0; i <= k; i++) {
			dfs(rt, i, 1, 0);
		}
		for (int i = 1; i <= n; i++) {
			ll ans = 0;
			for (int j = 0; j <= k; j++) {
				ans += ((sum1[i][j] + sum2[i][j]) % mod) * q_pow(deep[i], j) % mod;
				ans %= mod;
			}
			i == 1 ? printf("%lld", ans) : printf(" %lld", ans);
		}
		puts("");
	}
}