洛谷P2146 树链剖分

题意

思路：直接树链剖分，用线段树维护即可，算是树剖的经典题目吧。

代码：

#include <bits/stdc++.h>
#define ls(x) (x << 1)
#define rs(x) ((x << 1) | 1)
using namespace std;
const int maxn = 100010;
int head[maxn], Next[maxn * 2], ver[maxn * 2];
int sz[maxn], son[maxn], d[maxn], dfn[maxn], top[maxn], f[maxn];
int tot, cnt;
int n;
struct SegmentTree {
	int val, lz;
	int l, r;
};
SegmentTree tr[maxn * 4];
void add(int x, int y) {
	ver[++tot] = y;
	Next[tot] = head[x];
	head[x] = tot;
}
void dfs1(int x, int fa = -1) {
	sz[x] = 1;
	f[x] = fa;
	int mx = 0;
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa) continue;
		d[y] = d[x] + 1;
		dfs1(y, x);
		sz[x] += sz[y];
		if(sz[y] > mx) {
			mx = sz[y];
			son[x] = y;
		}
	}
}
void dfs2(int x, int fa, int t) {
	dfn[x] = ++cnt;
	top[x] = t;
	if(son[x]) dfs2(son[x], x, t);
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa || y == son[x]) continue;
		dfs2(y, x, y); 
	}	
}
void pushup(int o) {
	tr[o].val = tr[ls(o)].val + tr[rs(o)].val;
}
void maintain(int o, int val) {
	tr[o].val = val * (tr[o].r - tr[o].l + 1);
	tr[o].lz = val;
}
void pushdown(int o) {
	if(tr[o].lz != -1) {
		maintain(ls(o), tr[o].lz);
		maintain(rs(o), tr[o].lz);
		tr[o].lz = -1;
	}
}
void build(int o, int l, int r) {
	tr[o].l = l, tr[o].r = r;
	if(l == r) {
		tr[o].val = 0;
		tr[o].lz = -1;
		return;
	}
	int mid = (l + r) >> 1;
	build(ls(o), l, mid);
	build(rs(o), mid + 1, r);
	pushup(o);
}
void update(int o, int l, int r, int ql, int qr, int val) {
	if(l >= ql && r <= qr) {
		tr[o].val = (r - l + 1) * val;
		tr[o].lz = val;
		return;
	}
	pushdown(o);
	int mid = (l + r) >> 1;
	if(ql <= mid) update(ls(o), l, mid, ql, qr, val);
	if(qr > mid) update(rs(o), mid + 1, r, ql, qr, val);
	pushup(o);
}
int query(int o, int l, int r, int ql, int qr) {
	if(l >= ql && r <= qr) {
		return tr[o].val;
	}
	pushdown(o);
	int mid = (l + r) >> 1, ans = 0;
	if(ql <= mid) ans += query(ls(o), l, mid , ql, qr);
	if(qr > mid) ans += query(rs(o), mid + 1, r, ql, qr);
	return ans;
}
int solve(int x) {
	int ans = 0, st = x;
	while(x != -1) {
		ans += query(1, 1, n, dfn[top[x]], dfn[x]);
		x = f[top[x]];
	}
	return d[st] - d[0] + 1 - ans;
}
void update1(int x, int val) {
	while(x != -1) {
		update(1, 1, n, dfn[top[x]], dfn[x], val);
		x = f[top[x]];
	}
}
char s[110];
int main() {
	int x, m;
	scanf("%d", &n);
	for (int i = 1; i < n; i++)	{
		scanf("%d", &x);
		add(x, i);
		add(i, x);
	}
	f[0] = -1;
	build(1, 1, n);
	dfs1(0);
	dfs2(0, -1, 0);
	scanf("%d", &m);
	while(m--) {
		scanf("%s", s + 1);
		if(s[1] == 'i') {
			scanf("%d", &x);
			int tmp = query(1, 1, n, dfn[x], dfn[x]);
			if(tmp == 1) {
				printf("0
");
				continue;
			}
			printf("%d
", solve(x));
			update1(x, 1);
		} else {
			scanf("%d", &x);
			int tmp = query(1, 1, n, dfn[x], dfn[x]);
			if(tmp == 0) {
				printf("0
");
				continue;
			} 
			printf("%d
", query(1, 1, n, dfn[x], dfn[x] + sz[x] - 1));
			update(1, 1, n, dfn[x], dfn[x] + sz[x] - 1, 0);
		}
	}
}