Codeforces 877E Danil and a Part-time Job（dfs序 + 线段树）

题目链接   Danil and a Part-time Job

题意    给出一系列询问或者修改操作

　　　$pow$ $x$表示把以$x$为根的子树的所有结点的状态取反（$0$变$1$,$1$变$0$）

　　　$get$  $x$表示求以$x$为根的子树中状态为$1$的结点数。

首先大力$dfs$序，然后线段树操作一下。

具体问题转化为：区间翻转，区间求和。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define ls		i << 1
#define	rs		i << 1 | 1
#define lson		ls, L, mid
#define	rson		rs, mid + 1, R

const int N = 2e5 + 10;

struct node{
	int s[2];
	int len;
} t[N * 3];

int in[N], out[N], dc[N * 3], dt[N * 3], f[N], a[N];
int n, q, ti;

vector <int> v[N];

void dfs(int x){
	in[x] = ++ti;
	f[ti] = x;
	for (auto u : v[x]) dfs(u);
	out[x] = ti;
}

node update(node x, node y){
	node ans;
	ans.len = x.len + y.len;
	rep(i, 0, 1) ans.s[i] = x.s[i] + y.s[i];
	return ans;
}

void build(int i, int L, int R){
	if (L == R){
		int z = a[f[L]];
		t[i].len = t[i].s[z] = 1;
		t[i].s[z ^ 1] = 0;
		dc[i] = -1;
		dt[i] = 0;
		return ;
	}

	int mid = (L + R) >> 1;
	build(lson);
	build(rson);
	t[i] = update(t[ls], t[rs]);
	dc[i] = -1;
	dt[i] = 0;
}

void cover(int i, int z){
	dc[i] = z;
	dt[i] = 0;
	t[i].s[z] = t[i].len;
	t[i].s[z ^ 1] = 0;
}

void turn(int i, int z){
	if (~z){
		dc[i] ^= 1;
		t[i].s[z] = 0;
		t[i].s[z ^ 1] = t[i].len;
	}

	else{
		dt[i] ^= 1;
		swap(t[i].s[0], t[i].s[1]);
	}
}

void pushdown(int i){
	if (~dc[i]){
		cover(ls, dc[i]);
		cover(rs, dc[i]);
		dc[i] = -1;
	}

	if (dt[i]){
		turn(ls, dc[ls]);
		turn(rs, dc[rs]);
		dt[i] = 0;
	}
}

void Turn(int i, int L, int R, int x, int y){
	if (x <= L && R <= y){
		turn(i, dc[i]);
		return;
	}

	pushdown(i);
	int mid = (L + R) >> 1;
	if (x <= mid) Turn(lson, x, y);
	if (y > mid)  Turn(rson, x, y);
	t[i] = update(t[ls], t[rs]);
}

int query(int i, int L, int R, int x, int y){
	int ret = 0;
	if (x <= L && R <= y) return t[i].s[1];
	int mid = (L + R) >> 1;
	pushdown(i);
	if (x <= mid) ret += query(lson, x, y);
	if (y > mid ) ret += query(rson, x, y);
	return ret;
}

int main(){

	scanf("%d", &n);
	rep(i, 2, n){
		int x;
		scanf("%d", &x);
		v[x].push_back(i);
	}

	dfs(1);
	rep(i, 1, n) scanf("%d", a + i);
	build(1, 1, n);

	for (scanf("%d", &q); q--; ){
		char op[10]; int x;
		scanf("%s%d", op, &x);
		if (op[0] == 'g') printf("%d\n", query(1, 1, n, in[x], out[x]));
		else Turn(1, 1, n, in[x], out[x]);
	}

	return 0;
}