[bzoj4196][Noi2015]软件包管理器

题目大意:有一棵树,两个修改

  1. $install;x:$把根节点到$x$软件路径上的值全部变为$1$,并输出修改的节点个数
  2. $uninstall;x:$把$x$以及它的子树的值变为$0$,并输出修改的节点个数

题解:树链剖分,比较一下修改前后值的变化,即为答案

卡点:1.边忘记开两倍

C++ Code:

#include <cstdio>
#define maxn 100010
using namespace std;
int n, m;

int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxn << 1];
void add(int a, int b) {
	e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}

int dep[maxn], sz[maxn], fa[maxn];
int dfn[maxn], son[maxn], top[maxn], idx;
void dfs1(int rt) {
	sz[rt] = 1;
	for (int i = head[rt]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa[rt]) {
			fa[v] = rt;
			dep[v] = dep[rt] + 1;
			dfs1(v);
			sz[rt] += sz[v];
			if (!son[rt] || sz[son[rt]] < sz[v]) son[rt] = v;
		}
	}
}
void dfs2(int rt) {
	dfn[rt] = ++idx;
	int v = son[rt];
	if (v) top[v] = top[rt], dfs2(v);
	for (int i = head[rt]; i; i = e[i].nxt) {
		v = e[i].to;
		if (v != fa[rt] && v != son[rt]) {
			top[v] = v;
			dfs2(v);
		}
	}
}

int V[maxn << 2];
void add(int rt, int l, int r, int L, int R, int op) {
	if (L <= l && R >= r) {
		V[rt] = (r - l + 1) * op;
		return ;
	}
	if (V[rt] == r - l + 1) {
		V[rt << 1] = r - l + 2 >> 1;
		V[rt << 1 | 1] = r - l + 1 >> 1;
	}
	int mid = l + r >> 1;
	if (L <= mid) add(rt << 1, l, mid, L, R, op);
	if (R > mid) add(rt << 1 | 1, mid + 1, r, L, R, op);
	V[rt] = V[rt << 1] + V[rt << 1 | 1];
}
int ask(int rt, int l, int r, int L, int R) {
	if (L <= l && R >= r) return V[rt];
	if (V[rt] == r - l + 1) {
		V[rt << 1] = r - l + 2 >> 1;
		V[rt << 1 | 1] = r - l + 1 >> 1;
	}
	if (V[rt] == 0) V[rt << 1] = V[rt << 1 | 1] = 0;
	int mid = l + r >> 1, ans = 0;
	if (L <= mid) ans = ask(rt << 1, l, mid, L, R);
	if (R > mid) ans += ask(rt << 1 | 1, mid + 1, r, L, R);
	return ans;
}

inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
void modify(int x, int y, int num) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		add(1, 1, n, dfn[top[x]], dfn[x], num);
		x = fa[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	add(1, 1, n, dfn[x], dfn[y], num);
}
int query(int x, int y) {
	int ans = 0;
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		ans += ask(1, 1, n, dfn[top[x]], dfn[x]);
		x = fa[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	ans += ask(1, 1, n, dfn[x], dfn[y]);
	return ans;
}

int root = 1;
int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int a;
		scanf("%d", &a);
		add(a + root, i + root);
		add(i + root, a + root);
	}
	dep[top[root] = root] = 1;
	dfs1(root);
	dfs2(root);
	scanf("%d", &m);
	while (m --> 0) {
		char op[15];
		int ans, x;
		scanf("%s%d", op, &x); x += root;
		if (op[0] == 'i') {
			ans = query(root, x);
			modify(root, x, 1);
			printf("%d
", query(root, x) - ans);
		} else {
			ans = ask(1, 1, n, dfn[x], dfn[x] + sz[x] - 1);
			add(1, 1, n, dfn[x], dfn[x] + sz[x] - 1, 0);
			printf("%d
", ans - ask(1, 1, n, dfn[x], dfn[x] + sz[x] - 1));
		}
	}
	return 0;
}
原文地址:https://www.cnblogs.com/Memory-of-winter/p/9519699.html