【poj3321】Apple Tree——树状数组&DFS

题目

Apple Tree

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 24449 Accepted: 7295

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

3

1 2

1 3

3

Q 1

C 2

Q 1

Sample Output

3

2

Source

POJ Monthly–2007.08.05, Huang, Jinsong

描述：

给定一颗树，初始时每个节点上的权值均为1，之后有m次操作。

C x表示将编号为x的节点上的权值取反

Q x表示查询以x为根节点的子树的权值和

题解：

用树状数组来维护区间的权值和，用dfs序来维护每个节点对应的区间，dfs开始时的深度为该节点对应区间的下界，dfs返回时的深度为该节点对应区间的上界，注意在dfs时深度要用

++depth

来维护，否则的话如果该节点是叶节点则区间长度不是0而是1。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100005;
struct edge
{
int v,nxt;
}e[maxn];
int head[maxn],c[maxn],low[maxn],high[maxn],depth,n;
bool vis[maxn],hasApple[maxn];
int lowbit(int x)
{
return x & (-x);
}
int cal(int x)
{
int sum = 0;
for(;x;x -= lowbit(x))sum += c[x];
return sum;
}
void update(int x,int val)
{
for(;x <= n;x += lowbit(x))c[x] += val;
}
void dfs(int t)
{
low[t] = ++depth;
vis[t] = true;
for(int i = head[t];i > 0;i = e[i].nxt)
{
if(vis[e[i].v] == false)
{
dfs(e[i].v);
}
}
high[t] = depth;
}

int main()
{
//freopen("in.txt","r",stdin);
while(~scanf("%d",&n))
{
int u,v,cnt = 1;
for(int i = 0;i < n-1;i++)
{
scanf("%d%d",&u,&v);
e[cnt].v = v;
e[cnt].nxt = head[u];
head[u] = cnt++;
}
depth = 0;                                  //一定要从0开始++depth，因为不知道当前节点还有没有孩子，如果没有的话返回的depth就会比自身的depth多1
dfs(1);
for(int i = 1;i <= n;i++)
{
update(i,1);
hasApple[i] = true;
}
int m;scanf("%d",&m);
while(m--)
{
char op;
int x;
getchar();
scanf("%c %d",&op,&x);
if(op == 'C')
{
if(hasApple[x])
{
update(low[x], -1);
hasApple[x] = false;
}
else
{
update(low[x], 1);
hasApple[x] = true;
}
}
else
{
printf("%d\n",cal(high[x]) - cal(low[x] - 1));
}
}
}
return 0;
}