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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 27092 | Accepted: 8033 |
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
3 1 2 1 3 3 Q 1 C 2 Q 1
Sample Output
3 2
Source
1 /* 2 挺好的一道题，本题要重新建立树状数组，用dfs从树根开始搜，每搜到一个点他的左值记录该点的新序号，他的 3 右值记录改点的子树中的最大深度，这样每个点就变成了树状数组中的C数组参考上图，然后套树状数组区间求和的模板就行了。 4 注意本题如果用vector会很费时，vector<int>a[MAX]会超时，vector<vector<int> > a(MAX) 985ms过了(后面这个什么鬼!)。 6 不如自己建树用时会更少。 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<vector> 11 #include<cstring> 12 using namespace std; 13 const int MAX=100005; 14 int c[MAX]; 15 int lef[MAX],rig[MAX]; 16 int n,m; 17 int rot; 18 int head[MAX]; 19 int lne; 20 //vector<int>a[MAX]; 21 //vector<vector<int> > a(MAX); 22 struct node 23 { 24 int to,next; 25 }a[MAX]; 26 void tadd(int u,int v) 27 { 28 a[lne].to=v; 29 a[lne].next=head[u]; 30 head[u]=lne++; 31 } 32 int lowbit(int x) 33 { 34 return x&(-x); 35 } 36 void add(int id,int val) 37 { 38 while(id<=n) 39 { 40 c[id]+=val; 41 id+=lowbit(id); 42 } 43 } 44 int sum(int id) 45 { 46 int sum=0; 47 while(id>0) 48 { 49 sum+=c[id]; 50 id-=lowbit(id); 51 } 52 return sum; 53 } 54 void dfs(int x) 55 { 56 lef[x]=rot; 57 //for(int i=0;i<a[x].size();i++) 58 for(int i=head[x];i!=-1;i=a[i].next) 59 { 60 rot++; 61 //dfs(a[x][i]); 62 dfs(a[i].to); 63 } 64 rig[x]=rot; 65 } 66 int main() 67 { 68 int x,y; 69 char ch[5]; 70 int flag[MAX]; 71 scanf("%d",&n); 72 for(int i=1;i<=n;i++) 73 { 74 flag[i]=1; 75 add(i,1); 76 } 77 lne=0; 78 memset(head,-1,sizeof(head)); 79 for(int i=1;i<n;i++) 80 { 81 scanf("%d%d",&x,&y); 82 //a[x].push_back(y); 83 tadd(x,y); 84 } 85 rot=1; 86 dfs(1); 87 scanf("%d",&m); 88 while(m--) 89 { 90 scanf("%s%d",ch,&x); 91 if(ch[0]==‘C‘) 92 { 93 if(flag[x]) 94 add(lef[x],-1); 95 else add(lef[x],1); 96 flag[x]=!flag[x]; 97 } 98 else 99 { 100 printf("%d\n",sum(rig[x])-sum(lef[x]-1)); 101 } 102 } 103 return 0; 104 }
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原文地址：http://www.cnblogs.com/--ZHIYUAN/p/5935636.html
