3321 Apple Tree 树状数组

LIANJIE:http://poj.org/problem?id=3321

给你一个多叉树，每个叉和叶子节点有一颗苹果。然后给你两个操作，一个是给你C清除某节点上的苹果或者添加（此节点上有苹果则清除没苹果就添加）q询问这个节点的子树上有多少个苹果。

直接dfs遍历一遍，每个节点给一个时间戳，记录一下遍历玩这个节点是的始终时间戳，直接对时间戳进行询问即可。

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <stdlib.h>
 #include <vector>
 #include <queue>
 #include <stack>
 #define loop(s,i,n) for(i = s;i < n;i++)
 #define cl(a,b) memset(a,b,sizeof(a))
 #define lowbit(x) x&-x

 using namespace std;
 const int maxn = ;

 struct node
 {
     int u,v,w,next;
 }edges[maxn*];
 int head[maxn];
 bool vis[maxn];
 int dfsclock,n,cnt,pt[maxn],c[maxn];
 struct time
 {
     int s,e;
 }a[maxn];
 void init()
 {
     int i;
     for(i = ;i <= n;i++)
     head[i] = -,vis[i] = ,c[i] = ;
     cnt = dfsclock = ;
 }
 int sum(int x)
 {
     int ret = ;
     while(x > )
     {
         ret+=c[x];
         x-=lowbit(x);
     }
     return ret;
 }
 void add(int x,int d)
 {
     while(x <= n)
     {
         c[x]+= d;
         x+=lowbit(x);
     }
 }
 void addedge(int u,int v,int w)
 {
     int i;
     edges[cnt].u = u;
     edges[cnt].v = v;
     edges[cnt].w = w;
     edges[cnt].next = head[u];
     head[u] = cnt;
     cnt++;
 }
 void dfs(int u)
 {
     dfsclock++;
     vis[u] = ;
     a[u].s = dfsclock;
     int i;
     for(i = head[u];i != -;i = edges[i].next)
     {
         int v;
         v = edges[i].v;
         if(!vis[v])
         {
             dfs(v);
         }
     }
     a[u].e = dfsclock;

 }
 int main()
 {
     int q;
     while(~scanf("%d",&n))
     {

         int i;
         init();
         //memset(head,-1,sizeof(head));
         //cnt = 0;
         loop(,i,n-)
         {
             int u,v;
             scanf("%d %d",&u,&v);
             addedge(u,v,);
             addedge(v,u,);
         }

         dfs();

         for(i = ;i <= n;i++)
         add(i,);
         scanf("%d",&q);
         while(q--)
         {
             char s[];
             int num;
             scanf("%s%d",s,&num);
             if(s[] == 'Q')
             {
                 printf("%d\n",sum(a[num].e)-sum(a[num].s-));
             }
             else
             {
                 int leap;
                 leap = sum(a[num].s)-sum(a[num].s-);
               //  cout<<leap<<"**"<<endl;
                 if(leap)
                 add(a[num].s,-);
                 else
                 add(a[num].s,);
             }
         }

     }
     return ;
 }