Rokua P2982 [Usaco10feb] slow down slowing down (segment tree DFS sequence interval increment/decrement single point query)

To Los Valleys. 2982 slow down slowing down

Title Description

Every day Farmer John's n (1 <= N <= 100,000) cows conveniently numbered 1..N move from the barn to her priv Ate pasture. The pastures is organized as a tree, with the barn being on pasture 1. Exactly N-1 cow unidirectional paths connect the pastures; Directly connected pastures has exactly one path. Path I connects Pastures a_i and b_i (1 <= a_i <= N; 1 <= b_i <= N).

Cow I has a private pasture p_i (1 <= p_i <= N). The barn ' s small door lets only one cow exit at a time; And the patient cows wait until their predecessor arrives at her private pasture. First Cow 1 exits and moves to pasture p_1. Then Cow 2 exits and goes to pasture p_2, and so on.

While cow I walks to p_i she might or might don't pass through a pasture that already contains an eating cow. When a cow was present in a pasture, cow I walks slower than usual to prevent annoying her friend.

Consider the following pasture network, where the number betweenparentheses indicates the pastures' Owner.               1 (3)/(1) 4 3 (5)/\ (2) 2 5 (4) First, cow 1 walks to her pasture:1 (3) /[1] 3 (5)/\ (2) 2 5 (4) When Cow 2 moves to her pasture, she first passes into the barn ' spasture, Pastur E1.Then she sneaks around cow1In pasture4 beforearriving at her own pasture.1 (3)/[1]The3 (5)/\ [2]*5 (4) Cow3 doesn' t get far @ All--she lounges in the barn ' s pasture,#1.1* [3]/[1]The3 (5)/\ [2]*5 (4) Cow4 must slowFor pasture1and4On herTo Pasture5:1* [3]/[1]4* 3 (5)/\ [ 2] 2* 5* [ 4] Cow 5 slows for Cow 3 in pasture 1 and  Then enters her own private pasture: 1* [ 3]/[1] 4* 3*[5]/\ [2] 2* 5* [4]      

FJ would like to know what many times each cow have to slow down.

Farmer John's N-head cows every day (1 <= n <= 100000, number 1 ... N) went from the Granary to his own pasture. The pasture forms a tree, and the barn is on ranch 1th. There is a N-1 road directly connected to the pasture, so that there is a path between the pastures connected. The first road is connected to A_i,b_i, (1 <= a_i <= N; 1 <= b_i <= N). The cows each have a private ranch p_i (1 <= p_i <= N). The barn door can only let one cow leave at a time. The impatient cows will wait until their front friends arrive at their private ranch before leaving. First the cow 1 leaves, goes to p_1, then the Cow 2, and so on.

When cow I goes to pasture p_i, he may pass by a companion who is eating grass. When passing through a ranch that already has cows, cow I slows down its speed and prevents him from disturbing his friends.

FJ wanted to know how many times the cows slowed down in total.

Input/output format

Input format:

• Line 1:line 1 contains a single integer:n

• Lines 2..n:line i+1 contains, space-separated integers:a_i and B_i

• Lines n+1..n+n:line N+i contains a single integer:p_i

Output format:

• Lines 1..n:line I contains the number of times cow I have to slow down.

Input and Output Sample input example # #:

Sample # # of output:

Code:

1#include <cstdio>2#include <cctype>3 using namespacestd;4 Const intn=100005;5 6 intn,enum,cnt,h[n<<1],sum[n<<2],dfn[n],size[n];7 structEdge8 {9     intto,nxt;Ten}e[n<<1]; One  A voidReadint&Now ) - { -now=0;CharC=GetChar (); the      for(;! IsDigit (c); c=GetChar ()); -      for(; IsDigit (c); c=GetChar ()) -Now= (now<<3) + (now<<1) +c-'0'; - } +  - voidAddedge (intUintv) + { AE[++enum].to =v; atE[ENUM].NXT =H[u]; -H[u] =ENum; - } -  - voidDFS (intx) - { insize[x]=1;//The size of the root subtree (including itself) -dfn[x]=++cnt;//DFN[I]:I subscript in the DFS sequence to      for(intI=h[x];i;i=e[i].nxt) +     { -         intto=e[i].to; the         if(Dfn[to])Continue; * DFS (to); $size[x]+=Size[to];Panax Notoginseng     } - } the  + voidPushup (intRT) A { thesum[rt]=sum[rt<<1]+sum[rt<<1|1]; + } - voidPushdown (intRT) $ { $     if(! SUM[RT])return; -sum[rt<<1]+=Sum[rt]; -sum[rt<<1|1]+=Sum[rt]; thesum[rt]=0; - }Wuyi voidModifysum (intLintRintRtintLintR) the { -     if(L<=l && r<=R) Wu     { -++SUM[RT];//interval modification for lazy tag +1 About         return; $     } - pushdown (RT); -     intM= (l+r) >>1; -     if(l<=m) Modifysum (l,m,rt<<1, l,r); A     if(M<r) Modifysum (m+1,r,rt<<1|1, l,r); +     //pushup (RT); no pushup!! required the } - intQuerysum (intLintRintRtintp) $ { the     if(L==R)returnSum[rt]; the pushdown (RT); the     intM= (l+r) >>1, res=0; the     if(p<=m) Res+=querysum (l,m,rt<<1, p); -     ElseRes+=querysum (m+1,r,rt<<1|1, p); in     returnRes; the } the  About intMain () the { the read (n); the      for(intI=1; i<n;i++) +     { -         intb; the read (a); Read (b);Bayi Addedge (A, b); the Addedge (b,a); the     } -DFS (1); - //for (int i=1;i<=n;i++) the //printf ("%d:dfn:%d size:%d\n", I,dfn[i],size[i]); the      for(intI=1; i<=n;i++) the     { the         intA; - read (a); theprintf"%d\n", Querysum (1N1, Dfn[a]));//Go through and then change theModifysum (1N1, dfn[a],dfn[a]+size[a]-1);//single-point query interval modification the     }94     return 0; the}

Rokua P2982 [Usaco10feb] slow down slowing down (segment tree DFS sequence interval increment/decrement single point query)