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POJ 3321 Apple Tree

2018-06-17 21:38:07来源：未知阅读 ()

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 31499		Accepted: 9477

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

给你一颗树，支持两种操作

1.修改某一节点的权值

2.查询子树的权值（子树中节点的个数）

很显然可以用树状数组维护

为什么我的字体不能左对齐了。。。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<queue>
 6 #define lb(x) x&(-x)
 7 using namespace std;
 8 const int MAXN=2*1e6+10;
 9 const int INF=0x7fffff;
10 inline int read()
11 {
12     char c=getchar();int flag=1,x=0;
13     while(c<'0'||c>'9')    {if(c=='-')    flag=-1;c=getchar();}
14     while(c>='0'&&c<='9')    x=x*10+c-48,c=getchar();return x*flag;
15 }
16 struct node
17 {
18     int u,v,nxt;
19 }edge[MAXN];
20 int head[MAXN];
21 int num=1;
22 inline void add_edge(int x,int y)
23 {
24     edge[num].u=x;
25     edge[num].v=y;
26     edge[num].nxt=head[x];
27     head[x]=num++;
28 }
29 int n;
30 int in[MAXN],out[MAXN],cnt=0;
31 int tree[MAXN],have[MAXN];
32 inline void pre()
33 {
34     memset(head,-1,sizeof(head));
35     memset(have,1,sizeof(have));
36     num=1;cnt=0;
37 }
38 int deep[MAXN];
39 void dfs(int now)
40 {
41     in[now]=++cnt;
42     for(int i=head[now];i!=-1;i=edge[i].nxt)
43         dfs(edge[i].v);
44     out[now]=cnt;
45 }
46 int how[MAXN];//是否在树上 
47 inline void Interval_change(int pos,int val)
48 {
49     while(pos<=cnt)
50     {
51         tree[pos]+=val;
52         pos+=lb(pos);
53     }
54 }
55 inline int Interval_sum(int pos)
56 {
57     int ans=0;
58     while(pos)
59     {
60         ans+=tree[pos];
61         pos-=lb(pos);
62     }
63     return ans;
64 }
65 int main()
66 {
67     while(scanf("%d",&n)==1)
68     {
69         pre();
70         for(int i=1;i<=n-1;i++)
71         {
72             int x=read(),y=read();
73             add_edge(x,y);
74         }
75         dfs(1);
76         for(int i=1;i<=n;i++)
77             Interval_change(in[i],1);
78         int q=read();
79         for(int i=1;i<=q;i++)
80         {
81             char c[4];scanf("%s",c);
82             if(c[0]=='Q')//查询 
83             {
84                 int pos=read();
85                 printf("%d\n",Interval_sum(out[pos])-Interval_sum(in[pos]-1));
86             }
87             else
88             {
89                 int pos=read();
90                 if(have[pos])
91                     Interval_change(in[pos],-1);
92                 else Interval_change(in[pos],1);
93                 have[pos]=!have[pos];
94             }
95         }
96     }
97     return 0;
98 }