To 洛谷.2982 慢下來Slowing down

題目描述

Every day each of Farmer John‘s N (1 <= N <= 100,000) cows conveniently numbered 1..N move from the barn to her private pasture. The pastures are organized as a tree, with the barn being on pasture 1. Exactly N-1 cow unidirectional paths connect the pastures; directly connected pastures have 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 not pass through a pasture that already contains an eating cow. When a cow is present in a pasture, cow i walks slower than usual to prevent annoying her friend.

Consider the following pasture network, where the number between
parentheses indicates the pastures‘ owner.

        1 (3)        
       /   (1) 4   3 (5)
     / \   
(2) 2   5 (4)

First, cow 1 walks to her pasture:

        1 (3)        
       /   [1] 4*  3 (5)
     / \   
(2) 2   5 (4)

When cow 2 moves to her pasture, she first passes into the barn‘s
pasture, pasture 1. Then she sneaks around cow 1 in pasture 4 before
arriving at her own pasture.

        1 (3)
       /   [1] 4*  3 (5)
     / \   
[2] 2*  5 (4)

Cow 3 doesn‘t get far at all -- she lounges in the barn‘s pasture, #1.

        1* [3]
       /   [1] 4*  3 (5)
     / \   
[2] 2*  5 (4)

Cow 4 must slow for pasture 1 and 4 on her way to pasture 5:

        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 how many times each cow has to slow down.

每天Farmer John的N頭奶牛(1 <= N <= 100000，編號1…N)從糧倉走向他的自己的牧場。牧場構成了一棵樹，糧倉在1號牧場。恰好有N-1條道路直接連接著牧場，使得牧場之間都恰好有一條路徑相連。第i條路連接著A_i，B_i，(1 <= A_i <= N; 1 <= B_i <= N)。 奶牛們每人有一個私人牧場P_i (1 <= P_i <= N)。糧倉的門每次只能讓一只奶牛離開。耐心的奶牛們會等到他們的前面的朋友們到達了自己的私人牧場後才離開。首先奶牛1離開，前往P_1；然後是奶牛2，以此類推。

當奶牛i走向牧場P_i時候，他可能會經過正在吃草的同伴旁。當路過已經有奶牛的牧場時，奶牛i會放慢自己的速度，防止打擾他的朋友。

FJ想要知道奶牛們總共要放慢多少次速度。

輸入輸出格式

輸入格式：

• Line 1: Line 1 contains a single integer: N

• Lines 2..N: Line i+1 contains two space-separated integers: A_i and B_i

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

輸出格式：

• Lines 1..N: Line i contains the number of times cow i has to slow down.

輸入輸出樣例

5 
1 4 
5 4 
1 3 
2 4 
4 
2 
1 
5 
3 

0 
1 
0 
2 
1 

代碼：

 1 #include<cstdio>
 2 #include<cctype>
 3 using namespace std;
 4 const int N=100005;
 5 
 6 int n,ENum,cnt,H[N<<1],Sum[N<<2],Dfn[N],Size[N];
 7 struct Edge
 8 {
 9     int to,nxt;
10 }e[N<<1];
11 
12 void read(int &now)
13 {
14     now=0;char c=getchar();
15     for(;!isdigit(c);c=getchar());
16     for(;isdigit(c);c=getchar())
17       now=(now<<3)+(now<<1)+c-‘0‘;
18 }
19 
20 void AddEdge(int u,int v)
21 {
22     e[++ENum].to = v;
23     e[ENum].nxt = H[u];
24     H[u] = ENum;
25 }
26 
27 void DFS(int x)
28 {
29     Size[x]=1;//根的子樹的大小(包括自己) 
30     Dfn[x]=++cnt;//Dfn[i]:i在dfs序中的下標 
31     for(int i=H[x];i;i=e[i].nxt)
32     {
33         int to=e[i].to;
34         if(Dfn[to]) continue;
35         DFS(to);
36         Size[x]+=Size[to];
37     }
38 }
39 
40 void PushUp(int rt)
41 {
42     Sum[rt]=Sum[rt<<1]+Sum[rt<<1|1];
43 }
44 void PushDown(int rt)
45 {
46     if(!Sum[rt]) return;
47     Sum[rt<<1]+=Sum[rt];
48     Sum[rt<<1|1]+=Sum[rt];
49     Sum[rt]=0;
50 }
51 void ModifySum(int l,int r,int rt,int L,int R)
52 {
53     if(L<=l && r<=R)
54     {
55         ++Sum[rt];//區間修改時 針對本題 懶標記+1 
56         return;
57     }
58     PushDown(rt);
59     int m=(l+r)>>1;
60     if(L<=m) ModifySum(l,m,rt<<1,L,R);
61     if(m<R) ModifySum(m+1,r,rt<<1|1,L,R);
62     //PushUp(rt); 不需要進行PushUp!! 
63 }
64 int QuerySum(int l,int r,int rt,int p)
65 {
66     if(l==r) return Sum[rt];
67     PushDown(rt);
68     int m=(l+r)>>1,res=0;
69     if(p<=m) res+=QuerySum(l,m,rt<<1,p);
70     else res+=QuerySum(m+1,r,rt<<1|1,p);
71     return res;
72 }
73 
74 int main()
75 {
76     read(n);
77     for(int i=1;i<n;i++)
78     {
79         int a,b;
80         read(a);read(b);
81         AddEdge(a,b);
82         AddEdge(b,a);
83     }
84     DFS(1);
85 //    for(int i=1;i<=n;i++)
86 //      printf("%d:dfn:%d  size:%d\n",i,Dfn[i],Size[i]);
87     for(int i=1;i<=n;i++)
88     {
89         int a;
90         read(a);
91         printf("%d\n",QuerySum(1,n,1,Dfn[a]));//先經過後更改 
92         ModifySum(1,n,1,Dfn[a],Dfn[a]+Size[a]-1);//單點查詢 區間修改 
93     }
94     return 0;
95 }

洛谷P2982 [USACO10FEB]慢下來Slowing down(線段樹 DFS序 區間增減 單點查詢)