poj 3321 难度：一般

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Apple Tree

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 30842 Accepted: 9233

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

31 21 33Q 1C 2Q 1

Sample Output

32

题意：给你一颗苹果树，每个树杈有且只有一个苹果，注意不是二叉树是多叉树，Q: x表示x节点开始所有子树的苹果树之和，C:x表示如果x节点没有苹果，就加上一个苹果，如果右苹果，就减去一个苹果。

思路：看了别人的博客学会的  ： 点击打开链接

先记录每个节点所涵盖子树的区间范围，接下来就是树状数组区间求和的问题了，注意这题vector有点坑，代码有注释。

#include<cstdio>#include<vector>using namespace std;int left[200001],right[200001];int c[200000];int flag[200001];int tot,n;vector<vector<int> >G(100001);//vector<int>这种会超时，要用上面这种 void dfs(int u){left[u]=tot;//记录该节点的子树左区间 for(int i=0;i<G[u].size();i++){tot++;dfs(G[u][i]);}right[u]=tot;//记录该节点的子树右区间 }void add(int i,int key){while(i<=n){c[i]+=key;i+=i&(-i);}}int sum(int i){int ans=0;while(i>0){ans+=c[i];i-=i&(-i);}return ans;}int main(){int m;while(~scanf("%d",&n)){int i,u,v,t;char ch[2];for(i=0;i<=n;i++){G[i].clear();flag[i]=c[i]=0;}for(i=1;i<n;i++){scanf("%d%d",&u,&v);G[u].push_back(v);}tot=1;dfs(1);//计算每个节点的子树区间范围 for(i=1;i<=n;i++)add(i,1);scanf("%d",&m);while(m--){scanf("%s",ch);if(ch[0]=='Q'){scanf("%d",&t);printf("%d\n",sum(right[t])-sum(left[t]-1));//子树右区间的和-左区间的和 }else{scanf("%d",&t);if(!flag[t]){flag[t]=1;add(left[t],-1);//从该节点的子树做区间开始更新 }else{flag[t]=0;add(left[t],1);}}}}}