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INS14L.cpp
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// one of the best problems I've seen so far
// SPOJ : INS14L
#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define vi vector<ll>
#define pp pair<ll,ll>
#define mp make_pair
#define PI acos(-1.0)
#define all(v) v.begin(),v.end()
#define pb push_back
#define FOR(i,a,b) for(i=a;i<b;i++)
#define FREV(i,a,b) for(i=a;i>=b;i--)
#define SULL(n) scanf("%llu", &n)
#define INF 1e18
#define MOD 1000000007
#ifndef ONLINE_JUDGE
#define gc getchar
#define pc putchar
#else
#define gc getchar_unlocked
#define pc putchar_unlocked
#endif
using namespace std;
int read_int() {
char c = gc();
while((c < '0' || c > '9') && c != '-') c = gc();
int ret = 0, neg = 0;
if (c == '-') neg = 1, c = gc();
while(c >= '0' && c <= '9') {
ret = 10 * ret + c - 48;
c = gc();
}
return neg ? -ret : ret;
}
ll read_ll() {
char c = gc();
while((c < '0' || c > '9') && c != '-') c = gc();
ll ret = 0;
int neg = 0;
if (c == '-') neg = 1, c = gc();
while(c >= '0' && c <= '9') {
ret = 10 * ret + c - 48;
c = gc();
}
return neg ? -ret : ret;
}
struct node {
vi adj;
};
vi start_time(100005), end_time(100005), subtree_size(100005), depth(100005), cumulative_depth(100005), parent(100005);
vector<node> tree;
ll timer = 0;
void dfs(ll u) {
ll i, v, len = tree[u].adj.size();
start_time[u] = ++timer;
subtree_size[u] = 1;
cumulative_depth[u] = depth[u];
FOR(i,0,len) {
v = tree[u].adj[i];
depth[v] = depth[u] + 1;
dfs(v);
subtree_size[u] += subtree_size[v];
cumulative_depth[u] += cumulative_depth[v];
}
end_time[u] = timer;
}
ll range_query(ll* t, ll n, ll l, ll r) {
ll sum=0;
for(l+=n, r+=n; l<r; l>>=1, r>>=1) {
if(l & 1)
sum = (sum + t[l++])%MOD;
if(r & 1)
sum = (sum + t[--r])%MOD;
}
return sum;
}
void point_update(ll* t, ll n, ll index, ll key) {
index += n;
t[index] = (t[index] + key)%MOD;
for(; index>1; index>>=1) {
t[index>>1] = (t[index] + t[index^1])%MOD;
}
}
void range_update(ll* t, ll n, ll l, ll r, ll key) {
for(l+=n, r+=n; l<r; l>>=1, r>>=1) {
if(l & 1) {
t[l] = (t[l] + key)%MOD;
l++;
}
if(r & 1) {
--r;
t[r] = (t[r] + key)%MOD;
}
}
}
ll point_query(ll* t, ll n, ll index) {
ll ans = 0;
for(index+=n; index > 0; index>>=1) {
ans = (ans + t[index])%MOD;
}
return ans;
}
int main() {
ll i,j,t,u,n,q,x,k,total;
n = read_ll();
tree.resize(n+1);
ll seg_tree1[2*n+2], seg_tree2[2*n+2], seg_tree3[2*n+2];
FOR(i,0,2*n+2) {
seg_tree1[i] = seg_tree2[i] = seg_tree3[i] = 0;
}
parent[1] = 0;
FOR(i,2,n+1) {
j = read_ll();
tree[j].adj.pb(i);
parent[i] = j;
}
dfs(1);
// FOR(i,1,n+1) {
// printf("i = %lld st = %lld et = %lld sz = %lld dep = %lld cdep = %lld\n", i,start_time[i],end_time[i],subtree_size[i],depth[i],cumulative_depth[i]);
// }
q = read_ll();
FOR(i,0,q) {
t = read_ll();
u = read_ll();
if (t == 1) {
x = read_ll();
k = read_ll();
total = subtree_size[u] * x + k * (cumulative_depth[u] - depth[u] * subtree_size[u]);
// printf("calculated total = %lld\n", total);
range_update(seg_tree1, n+1, start_time[u], end_time[u]+1, (x - (depth[u]*k) + MOD)%MOD);
range_update(seg_tree2, n+1, start_time[u], end_time[u]+1, k);
point_update(seg_tree3, n+1, start_time[parent[u]], total);
// cout<<"==========================================================\n";
// FOR(j,1,2*n+2) {
// cout<<seg_tree1[j]<<" ";
// }
// cout<<"\n==========================================================\n";
// cout<<"==========================================================\n";
// FOR(j,1,2*n+2) {
// cout<<seg_tree2[j]<<" ";
// }
// cout<<"\n==========================================================\n";
// cout<<"==========================================================\n";
// FOR(j,1,2*n+2) {
// cout<<seg_tree3[j]<<" ";
// }
// cout<<"\n==========================================================\n";
}
else {
ll val1 = (point_query(seg_tree1, n+1, start_time[u]) * subtree_size[u])%MOD;
ll val2 = (point_query(seg_tree2, n+1, start_time[u]) * cumulative_depth[u])%MOD;
ll val3 = range_query(seg_tree3, n+1, start_time[u], end_time[u]+1);
// printf("val1 = %lld val2 = %lld val3 = %lld\n", val1, val2, val3);
total = ((val1 + val2)%MOD + val3)%MOD;
if (total < 0)
total += MOD;
printf("%lld\n", total);
}
}
return 0;
}