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POJ3321Apple Tree[树转序列 BIT]
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 26995 | Accepted: 8007 |
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
Sample Input
31 21 33Q 1C 2Q 1
Sample Output
32
Source
//// main.cpp// poj3211//// Created by Candy on 9/19/16.// Copyright © 2016 Candy. All rights reserved.//#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=1e5+5,M=1e5+5,INF=1e9;int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();} return x*f;}int n,m,x;char c[5];struct edge{ int v,ne;}e[N<<1];int cnt=0,h[N];inline void ins(int u,int v){ cnt++; e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt; cnt++; e[cnt].v=u;e[cnt].ne=h[v];h[v]=cnt;}int a[N],p=0,st[N],ed[N];void dfs(int u,int fa){ p++;st[u]=p; for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(v==fa) continue; dfs(v,u); } ed[u]=p;}int bit[N];inline int lowbit(int x){ return x&-x;}void buildBit(){ for(int i=1;i<=n;i++){ a[i]=1; bit[i]+=a[i]; if(i+lowbit(i)<=n) bit[i+lowbit(i)]+=bit[i]; }}inline void add(int x,int d){ while(x<=n){ bit[x]+=d;x+=lowbit(x); }}inline int sum(int x){ int ans=0; while(x>0){ ans+=bit[x];x-=lowbit(x); } return ans;}inline int query(int l,int r){ return sum(r)-sum(l-1);}int main(int argc, const char * argv[]) { n=read(); for(int i=1;i<=n-1;i++) ins(read(),read()); dfs(1,0); buildBit(); m=read(); //for(int i=1;i<=n;i++) printf("%d %d %d %d\n",st[i],ed[i],a[i],bit[i]); for(int i=1;i<=m;i++){ scanf("%s",c); if(c[0]==‘C‘){ x=read();int b=st[x]; if(a[b]) add(b,-1); else add(b,1); a[b]^=1; } if(c[0]==‘Q‘){ x=read(); printf("%d\n",query(st[x],ed[x])); } } return 0;}
POJ3321Apple Tree[树转序列 BIT]