【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("");
}
}