There is a rooted tree with n nodes, number from 1-n. Root’s number is 1.Each node has a value ai.
Initially all the node’s value is 0.
We have q operations. There are two kinds of operations.
1 v x k : a[v]+=x , a[v’]+=x-k (v’ is child of v) , a[v’’]+=x-2*k (v’’ is child of v’) and so on.
2 v : Output a[v] mod 1000000007(10^9 + 7).
Input
First line contains an integer T (1 ≤ T ≤ 3), represents there are T test cases.
In each test case:
The first line contains a number n.
The second line contains n-1 number, p2,p3,…,pn . pi is the father of i.
The third line contains a number q.
Next q lines, each line contains an operation. (“1 v x k” or “2 v”)
1 ≤ n ≤ 3*10^5
1 ≤ pi < i
1 ≤ q ≤ 3*10^5
1 ≤ v ≤ n; 0 ≤ x < 10^9 + 7; 0 ≤ k < 10^9 + 7
Output
For each operation 2, outputs the answer.
Sample Input
1 3 1 1 3 1 1 2 1 2 1 2 2
Sample Output
2 1
题意
节点u加x，节点u的儿子u′加x−k，节点u的孙子u′′加x−2×k，依次类推。
看数据范围标准的线段树
但是这个k不好处理 这个k和深度有关系
所以这题在初始化处理的时候 1LL * x + 1LL * dep[v]*k
这样就便于我们查询了
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 #include <iostream> 8 #include <map> 9 #include <stack> 10 #include <string> 11 #include <vector> 12 #define rtl rt<<1 13 #define rtr rt<<1|1 14 #define mem(a,b) memset(a,b,sizeof(a)) 15 #define fuck(x) cout<<"["<<x<<"]"<<endl 16 #define sf(n) scanf("%d", &n) 17 #define sff(a,b) scanf("%d %d", &a, &b) 18 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) 19 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d) 20 #define FIN freopen("DATA.txt","r",stdin) 21 #define read(x) scanf("%d", &x) 22 using namespace std; 23 typedef long long LL; 24 const int maxn = 1e6 + 10; 25 const int mod = 1e9 + 7; 26 int t, n, q, tot, dfscnt, head[maxn], L[maxn], R[maxn], dep[maxn]; 27 struct Edge { 28 int v, nxt; 29 } edge[maxn << 1]; 30 void add(int u, int v) { 31 edge[tot].v = v; 32 edge[tot].nxt = head[u]; 33 head[u] = tot++; 34 } 35 void dfs(int u, int fa, int d) { 36 L[u] = ++dfscnt; 37 dep[u] = d; 38 for (int i = head[u] ; ~i ; i = edge[i].nxt) { 39 int v = edge[i].v; 40 if (v != fa) dfs(v, u, d + 1); 41 } 42 R[u] = dfscnt; 43 } 44 struct node { 45 int l, r; 46 LL num, k; 47 int mid() { 48 return (l + r) >> 1; 49 } 50 } tree[maxn << 2]; 51 void build(int l, int r, int rt) { 52 tree[rt].l = l, tree[rt].r = r; 53 tree[rt].num = tree[rt].k = 0; 54 if (l == r) return ; 55 int m = (l + r) >> 1; 56 build(l, m, rtl); 57 build(m + 1, r, rtr); 58 } 59 void pushdown(int rt) { 60 if (tree[rt].num != 0) { 61 tree[rtl].num = (tree[rt].num + tree[rtl].num) % mod; 62 tree[rtr].num = (tree[rt].num + tree[rtr].num) % mod; 63 tree[rt].num = 0; 64 } 65 if (tree[rt].k != 0) { 66 tree[rtl].k = (tree[rt].k + tree[rtl].k) % mod; 67 tree[rtr].k = (tree[rt].k + tree[rtr].k) % mod; 68 tree[rt].k = 0; 69 } 70 } 71 72 void update(LL a, LL b, int L, int R, int rt) { 73 if (L <= tree[rt].l && tree[rt].r <= R) { 74 tree[rt].num = ((tree[rt].num + a) % mod + mod) % mod; 75 tree[rt].k = ((tree[rt].k + b) % mod + mod) % mod; 76 return ; 77 } 78 pushdown(rt); 79 int mid = tree[rt].mid(); 80 if (R <= mid) update(a, b, L, R, rtl); 81 else if(L > mid) update(a, b, L, R, rtr); 82 else { 83 update(a, b, L, mid, rtl); 84 update(a, b, mid + 1, R, rtr); 85 } 86 } 87 LL query(int pos, int d, int rt) { 88 if (tree[rt].l == tree[rt].r) { 89 return ((tree[rt].num - d * tree[rt].k % mod) % mod + mod) % mod; 90 } 91 pushdown(rt); 92 int mid = tree[rt].mid(); 93 if (pos <= mid) return query(pos, d, rtl); 94 return query(pos, d, rtr); 95 } 96 int main() { 97 int op, v, x, k; 98 sf(t); 99 while(t--) { 100 sf(n); 101 for (int i = 0 ; i <= n ; i++) head[i] = -1; 102 tot = 0; 103 for (int i = 2 ; i <= n ; i++) { 104 sf(v); 105 add(v, i); 106 add(i, v); 107 } 108 dfscnt = 0; 109 dfs(1, -1, 1); 110 build(1, n, 1); 111 sf(q); 112 while(q--) { 113 sf(op); 114 if (op == 1) { 115 sfff(v, x, k); 116 update(1LL * x + 1LL * dep[v]*k, k, L[v], R[v], 1); 117 } else { 118 sf(v); 119 printf("%lld ", query(L[v], dep[v], 1)); 120 } 121 } 122 } 123 return 0; 124 }