poj 3321 Ultra-QuickSort （树状数组）

Apple Tree

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 30850 Accepted: 9235

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

Source

POJ Monthly--2007.08.05, Huang, Jinsong

题意： 有一个苹果树，有两种操作，Q是询问所有子树包括结点一共有多少个苹果，C是如果结点存在苹果就吃掉，否则长出一个苹果。

分析： 额。。其实就是修改一下结点的编号，有个小坑，这个题居然卡 vector ~~所以要用邻接表做

AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<vector>using namespace std;const int maxn=1e5+20;int left[maxn],right[maxn];int sum[maxn];int head[maxn];bool s[maxn];int cnt;int num;struct tree{int u,v,next;tree(){}tree(int uu,int vv,int nn){u=uu,v=vv,next=nn;}}edge[maxn];int lowbit(int x){return x&(-x);}void init(int n){cnt=1;num=0;memset(sum,0,sizeof(sum));memset(s,false,sizeof(s));memset(head,-1,sizeof(head));}void addedge(int a,int b){edge[num].u=a;edge[num].v=b;edge[num].next=head[a];head[a]=num++;}void dfs(int x){left[x]=cnt;for(int i=head[x];i!=-1;i=edge[i].next){cnt++;dfs(edge[i].v);}right[x]=cnt;}void update(int x,int d,int n){while(x<=n){sum[x]+=d;x+=lowbit(x);}}int getadd(int x){int ans=0;while(x>0){ans+=sum[x];x-=lowbit(x);}return ans;}int main(){int n;while(scanf("%d",&n)==1){init(n);for(int i=0;i<n-1;i++){int a,b;scanf("%d%d",&a,&b);addedge(a,b);}dfs(1);for(int i=1;i<=n;i++){s[i]=true;update(i,1,n);}int m;scanf("%d",&m);for(int i=0;i<m;i++){char q[3];int w;scanf("%s",q);scanf("%d",&w);if(q[0]=='Q'){printf("%d\n",getadd(right[w])-getadd(left[w]-1));}else{if(s[w])update(left[w],-1,n);elseupdate(left[w],1,n);s[w]=!s[w];}}}}