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Initially all vertices contain number 0. Then come q queries, each query has one of the two types:
Process the queries given in the input.
The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn (1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.
The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.
For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).
3
1 1
3
1 1 2 1
2 1
2 2
2
1
You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 300010; 5 const int mod = 1000000007; 6 vector<int>g[maxn]; 7 LL c[2][maxn],val[2]; 8 int n,m,L[maxn],R[maxn],d[maxn],clk; 9 void update(int i){ 10 while(i < maxn){ 11 c[0][i] += val[0]; 12 c[1][i] += val[1]; 13 c[0][i] %= mod; 14 c[1][i] %= mod; 15 i += i&-i; 16 } 17 } 18 LL query(int i){ 19 LL sum[2] = {0},dep = d[i]; 20 i = L[i]; 21 while(i > 0){ 22 sum[0] += c[0][i]; 23 sum[1] += c[1][i]; 24 sum[0] %= mod; 25 sum[1] %= mod; 26 i -= i&-i; 27 } 28 return ((sum[0] - dep*sum[1])%mod + mod)%mod; 29 } 30 void dfs(int u,int dep){ 31 L[u] = ++clk; 32 d[u] = dep; 33 for(int i = g[u].size()-1; i >= 0; --i) 34 dfs(g[u][i],dep+1); 35 R[u] = clk; 36 } 37 int main(){ 38 int u,op,x,y,z; 39 while(~scanf("%d",&n)){ 40 for(int i = clk = 0; i <= n; ++i) g[i].clear(); 41 for(int i = 2; i <= n; ++i){ 42 scanf("%d",&u); 43 g[u].push_back(i); 44 } 45 dfs(1,0); 46 memset(c,0,sizeof c); 47 scanf("%d",&m); 48 while(m--){ 49 scanf("%d%d",&op,&x); 50 if(op == 1){ 51 scanf("%d%d",&y,&z); 52 val[0] = ((LL)y + (LL)d[x]*z)%mod; 53 val[1] = z; 54 update(L[x]); 55 val[0] = -val[0]; 56 val[1] = -val[1]; 57 update(R[x]+1); 58 }else printf("%I64d\n",query(x)); 59 } 60 } 61 return 0; 62 }
CodeForces 396C On Changing Tree
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原文地址：http://www.cnblogs.com/crackpotisback/p/4785056.html
