POJ - 3321Apple Tree
| Time Limit: 2000MS |
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Memory Limit: 65536KB |
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64bit IO Format: %I64d & %I64u |
Submit Status
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
/*Author: 2486Memory: 10004 KBTime: 766 MSLanguage: G++Result: Accepted*/#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <queue>using namespace std;#define lson rt << 1, l, mid#define rson rt << 1|1, mid + 1, r#define root 1, 1, Nconst int MAXN = 2e5 + 5;int sum[MAXN << 2], LU[MAXN], RU[MAXN], N, M, A, B, tot, K;char op[10];/***********加邊模板***************/int Head[MAXN], Next[MAXN], rear;struct edge { int u,v;} es[MAXN];void Edge_Init() { rear = 0; memset(Head, -1, sizeof(Head));}void Edge_Add(int u,int v) { es[rear].u = u; es[rear].v = v; Next[rear] = Head[u]; Head[u] = rear ++;}void DFS(int to, int from) { LU[to] = ++ tot;//用來標記屬於它的子樹的序列 for(int i = Head[to] ; ~ i; i = Next[i]) { int v = es[i].v; if(v == from) continue; DFS(v, to); } RU[to] = tot;}/**********************************/void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1|1];}void build(int rt, int l, int r) { if(l == r) { sum[rt] = 1; return ; } int mid = (l + r) >> 1; build(lson); build(rson); pushup(rt);}void update(int p,int rt,int l, int r) { if(l == r) { sum[rt] ^= 1; return; } int mid = (l + r) >> 1; if(p <= mid) update(p, lson); else update(p, rson); pushup(rt);}int query(int L, int R,int rt, int l, int r) { if(L <= l && r <= R) { return sum[rt]; } int mid = (l + r) >> 1; int res = 0; if(L <= mid) res += query(L, R, lson); if(R > mid) res += query(L, R, rson); return res;}int main() { //freopen("D://imput.txt","r",stdin); while(~ scanf("%d", &N)) { tot = 0; build(root); Edge_Init(); for(int i = 1; i < N ; i ++) { scanf("%d%d", &A, &B); Edge_Add(A, B); Edge_Add(B, A); } DFS(1, -1); scanf("%d", &M); while(M --) { scanf("%s %d", op, &K); if(op[0] == ‘C‘) { update(LU[K],root); } else { printf("%d\n", query(LU[K],RU[K],root)); } } } return 0;}
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