[poj3321]Apple Tree(dfs序+樹狀數組) - Pumbit-Legion－IT工程師數位筆記本

文章出處

Apple Tree

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26762		Accepted: 7947

Description

There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple tree.

The tree has N forks which are connected by branches. Kaka numbers the forks by 1 to N and the root is always numbered by 1. Apples will grow on the forks and two apple won't grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.

The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?

Input

The first line contains an integer N (N ≤ 100,000) , which is the number of the forks in the tree.
The following N - 1 lines each contain two integers u and v, which means fork u and fork v are connected by a branch.
The next line contains an integer M (M ≤ 100,000).
The following M lines each contain a message which is either
"C x" which means the existence of the apple on fork x has been changed. i.e. if there is an apple on the fork, then Kaka pick it; otherwise a new apple has grown on the empty fork.
or
"Q x" which means an inquiry for the number of apples in the sub-tree above the fork x, including the apple (if exists) on the fork x
Note the tree is full of apples at the beginning

Output

For every inquiry, output the correspond answer per line.

Sample Input

Sample Output

3
2

Source

POJ Monthly--2007.08.05, Huang, Jinsong

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dfs序嘛

原來之前理解的是錯的

其實就是tarjan的dfn與low數組

得出dfn與low后

dfn[i]是起點，low[i]是終點，中間的一段就是點i的子樹了

這樣dfs序就可以“方便的維護子樹”

而對于大部分按照dfs序維護序列的題，與樹狀數組是相當好的組合

這道題差不多是裸題了，多開一個表示當前節點狀態的數組就可以了

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 int ap[100010],bit[100010];
 5 int bg[100010],ed[100010],cnt=0;
 6 int n;
 7 typedef struct{
 8     int to,nxt;
 9 }edge;
10 edge gra[200010];
11 int head[100010],num=0;
12 int add(int frm,int to){
13     gra[++num].nxt=head[frm];
14     gra[num].to=to;
15     head[frm]=num;
16     return 0;
17 }
18 int dfs(int u,int fa){
19     bg[u]=++cnt;
20     int j;
21     for(j=head[u];j;j=gra[j].nxt){
22         if(gra[j].to!=fa)dfs(gra[j].to,u);
23     }
24     ed[u]=cnt;
25     return 0;
26 }
27 int lb(int x){
28     return x&(-x);
29 }
30 int c(int x){
31     int num=ap[x];
32     x=bg[x];
33     while(x<=n){
34         bit[x]+=num;
35         x+=lb(x);
36     }
37     return 0;
38 }
39 int q(int x){
40     int a1=0,a2=0;
41     int e=ed[x];
42     while(e){
43         a1+=bit[e];
44         e-=lb(e);
45     }
46     int b=bg[x]-1;
47     while(b){
48         a2+=bit[b];
49         b-=lb(b);
50     }
51     return a1-a2;
52 }
53 int main(){
54     scanf("%d",&n);
55     for(int i=1;i<n;i++){
56         int x,y;
57         scanf("%d %d",&x,&y);
58         add(x,y);
59         add(y,x);
60     }
61     dfs(1,0);
62     for(int i=1;i<=n;i++){
63         ap[i]=1;
64         c(i);
65     }
66     int m;
67     scanf("%d",&m);
68     for(int i=1;i<=m;i++){
69         char in[2];
70         int x;
71         scanf("%s %d",in,&x);
72         if(in[0]=='C'){
73             ap[x]*=-1;
74             c(x);
75         }
76         else printf("%d\n",q(x));
77     }
78     return 0;
79 }