多次修改一棵树节点的值，或者询问当前这个节点的子树所有节点权值总和。
首先预处理出DFS序L[i]和R[i]
把问题转化为区间查询总和问题。单点修改，区间查询，树状数组即可。
注意修改的时候也要按照dfs序修改，因为你查询就是按照dfs查的，所以修改也要用dfs序修改
L[i]是唯一的。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 100000 + 20;
int first[maxn], L[maxn], R[maxn], a[maxn];
struct edge {
int u, v;
int next;
} e[maxn * 2];
int n, num;
void add (int u, int v)
{
++num;
e[num].u = u;
e[num].v = v;
e[num].next = first[u];
first[u] = num;
}
bool book[maxn];
int index;
void dfs (int cur)
{
L[cur] = index;
for (int i = first[cur]; i; i = e[i].next) {
if (book[e[i].v] == 0) {
book[e[i].v] = 1;
index++;
dfs (e[i].v);
}
}
R[cur] = index;
}
int c[maxn];//树状数组，多case的记得要清空
int lowbit (int x)//得到x二进制末尾0的个数的2次方 2^num
{
return x&(-x);
}
void addc (int pos,int val)//在第pos位加上val这个值
{
while (pos<=n) { //n是元素的个数
c[pos] += val;
pos += lowbit(pos);
}
return ;
}
int get_sum (int pos) //求解：1--pos的总和
{
int ans = 0;
while (pos) {
ans += c[pos];
pos -= lowbit(pos);
}
return ans;
}
void work ()
{
for (int i = 1; i <= n - 1; ++i) {
int u, v;
scanf ("%d%d", &u, &v);
add (u, v);
}
for (int i = 1; i <= n; ++i) {
addc (i, 1);
a[i] = 1;
}
dfs (1);
int m;
scanf ("%d", &m);
for (int i = 1; i <= m; ++i) {
char str[3];
int id;
scanf ("%s%d", str, &id);
if (str[0] == 'Q') {
printf ("%d\n", get_sum (R[id]) - get_sum (L[id] - 1));
} else {
id = L[id]; //查询用dfs序查，那么更改也要用dfs序更改
a[id] = -a[id];
addc (id, a[id]);
}
}
return ;
}
int main ()
{
#ifdef local
freopen("data.txt","r",stdin);
#endif
while (~scanf ("%d", &n)) {
index = 1;
work ();
memset (first, 0, sizeof first);
num = 0;
memset (book, 0, sizeof book);
memset (c, 0, sizeof c);
}
return 0;
}
View Code
