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; }
时间: 2024-12-28 01:19:59