PKU 3321 - Apple Tree (树状数组 + DFS)
题意
有一颗苹果树,1是树根.
有两种操作
-
改变一个结点的值
-
统计这个结点以上(包括)的果实的个数。
思路
学习了时间戳。
就是在访问到一个结点的时候,放一个标记,等到递归结束的时候再放一个,这个范围就是他的子树的范围,然后用树状数组统计。真是。。太巧妙了。
这题的数组版邻接表和vector版相差也很大。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 2e5 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct BIT
{
int l, r, val;
}bit[MAXN];
struct EDGE
{
int from, to;
};
int lev, C[MAXN], head[MAXN], next[MAXN], vis[MAXN];
vector<EDGE> edges;
void AddEdge(int from, int to)
{
edges.PB((EDGE){from, to});
int k = SZ(edges) - 1;
next[k] = head[from];
head[from] = k;
}
void DFS(int x)
{
vis[x] = 1;
bit[x].l = ++lev;
for (int i = head[x]; i != -1; i = next[i])
if (!vis[edges[i].to]) DFS(edges[i].to);
bit[x].r = lev;
}
void Update(int x, int k)
{
for (int i = x; i <= lev; i += Lowbit(i))
if (!k) C[i]--;
else C[i]++;
}
int Query(int x)
{
int sum = 0;
while (x > 0)
{
sum += C[x];
x -= Lowbit(x);
}
return sum;
}
int main()
{
//ROP;
int n, i, j;
while (~scanf("%d", &n))
{
MS(head, -1); MS(C, 0); MS(vis, 0);
lev = 0;
for (i = 1; i <= n - 1; i++)
{
int u, v;
scanf("%d%d", &u, &v);
AddEdge(u, v);
AddEdge(v, u);
}
DFS(1);
for (i = 1; i <= n; i++)
{
Update(bit[i].l, 1);
bit[i].val = 1;
}
scanf("%d", &n);
while (n--)
{
char str[3];
int x;
scanf("%s%d", str, &x);
if (str[0] == 'Q') printf("%d\n", Query(bit[x].r) - Query(bit[x].l - 1));
else
{
bit[x].val ^= 1;
Update(bit[x].l, bit[x].val);
}
}
}
return 0;
}