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。
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <stdlib.h> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #define loop(s,i,n) for(i = s;i < n;i++) 10 #define cl(a,b) memset(a,b,sizeof(a)) 11 #define lowbit(x) x&-x 12 13 using namespace std; 14 const int maxn = 100005; 15 16 struct node 17 { 18 int u,v,w,next; 19 }edges[maxn*2]; 20 int head[maxn]; 21 bool vis[maxn]; 22 int dfsclock,n,cnt,pt[maxn],c[maxn]; 23 struct time 24 { 25 int s,e; 26 }a[maxn]; 27 void init() 28 { 29 int i; 30 for(i = 0;i <= n;i++) 31 head[i] = -1,vis[i] = 0,c[i] = 0; 32 cnt = dfsclock = 0; 33 } 34 int sum(int x) 35 { 36 int ret = 0; 37 while(x > 0) 38 { 39 ret+=c[x]; 40 x-=lowbit(x); 41 } 42 return ret; 43 } 44 void add(int x,int d) 45 { 46 while(x <= n) 47 { 48 c[x]+= d; 49 x+=lowbit(x); 50 } 51 } 52 void addedge(int u,int v,int w) 53 { 54 int i; 55 edges[cnt].u = u; 56 edges[cnt].v = v; 57 edges[cnt].w = w; 58 edges[cnt].next = head[u]; 59 head[u] = cnt; 60 cnt++; 61 } 62 void dfs(int u) 63 { 64 dfsclock++; 65 vis[u] = 1; 66 a[u].s = dfsclock; 67 int i; 68 for(i = head[u];i != -1;i = edges[i].next) 69 { 70 int v; 71 v = edges[i].v; 72 if(!vis[v]) 73 { 74 dfs(v); 75 } 76 } 77 a[u].e = dfsclock; 78 79 } 80 int main() 81 { 82 int q; 83 while(~scanf("%d",&n)) 84 { 85 86 int i; 87 init(); 88 //memset(head,-1,sizeof(head)); 89 //cnt = 0; 90 loop(0,i,n-1) 91 { 92 int u,v; 93 scanf("%d %d",&u,&v); 94 addedge(u,v,1); 95 addedge(v,u,1); 96 } 97 98 dfs(1); 99 100 for(i = 1;i <= n;i++) 101 add(i,1); 102 scanf("%d",&q); 103 while(q--) 104 { 105 char s[2]; 106 int num; 107 scanf("%s%d",s,&num); 108 if(s[0] == 'Q') 109 { 110 printf("%d ",sum(a[num].e)-sum(a[num].s-1)); 111 } 112 else 113 { 114 int leap; 115 leap = sum(a[num].s)-sum(a[num].s-1); 116 // cout<<leap<<"**"<<endl; 117 if(leap) 118 add(a[num].s,-1); 119 else 120 add(a[num].s,1); 121 } 122 } 123 124 } 125 return 0; 126 }