Time Limit: 2000 mSec Memory Limit : 262144 KB
Problem Description
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
Sample Output
Source
第八届福建省大学生程序设计竞赛-重现赛(感谢承办方厦门理工学院)lld wa一天
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define LL __int64 #define pi (4*atan(1.0)) #define eps 1e-8 #define bug(x) cout<<"bug"<<x<<endl; const int N=3e5+10,M=2e6+10,inf=1e9+10; const LL INF=1e18+10,mod=1e9+7; struct isss { int v,nex; }edge[N<<1]; int head[N],edg; int in[N],out[N],tot; LL deep[N]; void add(int u,int v) { ++edg; edge[edg].v=v; edge[edg].nex=head[u]; head[u]=edg; } void dfs(int u,int fa,int dp) { in[u]=++tot; deep[u]=dp; for(int i=head[u];i;i=edge[i].nex) { int v=edge[i].v; if(v==fa)continue; dfs(v,u,dp+1); } out[u]=tot; } struct AYT { LL tree[N]; void init() { memset(tree,0,sizeof(tree)); } int lowbit(int x) { return x&-x; } void update(int x,LL c) { while(x<N) { tree[x]+=c; x+=lowbit(x); } } LL query(int x) { LL ans=0; while(x) { ans+=tree[x]; x-=lowbit(x); } return ans; } }TX,TK; void init() { tot=0; memset(head,0,sizeof(head)); memset(deep,0,sizeof(deep)); TX.init(); TK.init(); edg=0; } int main() { int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); init(); for(int i=2;i<=n;i++) { int f; scanf("%d",&f); add(f,i); add(i,f); } dfs(1,-1,0); int q; scanf("%d",&q); while(q--) { int t,v; LL x,k; scanf("%d%d",&t,&v); if(t==1) { scanf("%I64d%I64d",&x,&k); x+=1LL*deep[v]*k; x%=mod; TX.update(in[v],x); TX.update(out[v]+1,-x); TK.update(in[v],k); TK.update(out[v]+1,-k); } else { LL xx=TX.query(in[v]); LL y=TK.query(in[v]); y%=mod; LL ans=xx-1LL*deep[v]*y; ans%=mod;ans+=mod;ans%=mod; printf("%I64d ",ans); } } } return 0; }