P2146 [NOI2015] 软件包管理器

思路分析

树链剖分板子，每次安装时，将 \(1\) 到 \(x\) 的链变为 \(1\)，卸载时，将 \(x\) 的子树变为 \(0\)。

代码

#include<iostream>

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 1e5 + 10;
int n, q;
struct edge{
	int v, nxt;
}e[N << 1];
int head[N], cnt;
int sz[N], dep[N], son[N], f[N], id[N], top[N], tot;
void add(int u, int v){
	e[++cnt] = (edge){v, head[u]};
	head[u] = cnt;
}
void dfs1(int u, int fa){
	sz[u] = 1, f[u] = fa, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == fa) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[v] > sz[son[u]]) son[u] = v;
	}
}
void dfs2(int u, int t){
	id[u] = ++tot;
	top[u] = t;
	if (son[u]) dfs2(son[u], t);
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == f[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
struct tree{
	int sum, tag;
}t[N << 2];
void pushup(int now){
	t[now].sum = t[now << 1].sum + t[now << 1 | 1].sum;
}
void pushdown(int now, int l, int r){
	if (t[now].tag != -1){
		int mid = (l + r) >> 1;
		t[now << 1].sum = t[now].tag * (mid - l + 1);
		t[now << 1 | 1].sum = t[now].tag * (r - mid);
		t[now << 1].tag = t[now].tag;
		t[now << 1 | 1].tag = t[now].tag;
		t[now].tag = -1;
	}
}
void build(int now, int l, int r){
	t[now].tag = -1, t[now].sum = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(now << 1, l, mid);
	build(now << 1 | 1, mid + 1, r);
}
void modify(int now, int l, int r, int x, int y, int k){
	if (x <= l && r <= y){
		t[now].sum = k * (r - l + 1), t[now].tag = k;
		return;
	}
	pushdown(now, l, r);
	int mid = (l + r) >> 1;
	if (x <= mid) modify(now << 1, l, mid, x, y, k);
	if (mid + 1 <= y) modify(now << 1 | 1, mid + 1, r, x, y, k);
	pushup(now);
}
void modify_chain(int x, int y, int k){
	int fx = top[x], fy = top[y];
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		modify(1, 1, n, id[fx], id[x], k);
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	modify(1, 1, n, id[x], id[y], k);
}

int main(){
	n = read();
	for (int i = 1; i < n; i++){
		int x = read() + 1;
		add(x, i + 1);
	}
	dfs1(1, 0);
	dfs2(1, 1);
	build(1, 1, n);
	q = read();
	for (int i = 1; i <= q; i++){
		char c[10];
		cin >> c;
		int OP = t[1].sum, x = read() + 1;
		if (c[0] == 'i'){
			modify_chain(1, x, 1);
			cout << t[1].sum - OP << '\n';
		}else{
			modify(1, 1, n, id[x], id[x] + sz[x] - 1, 0);
			cout << OP - t[1].sum << '\n';
		}
	}
	return 0;
} 

P2486 [SDOI2011]

思路分析

很好的树链剖分题，在线段树中记录区间最左端颜色和最右端颜色，合并即可。
注意，链与链之间的查询也需要判断合并中间的颜色是否相同。

代码

#include<iostream>
#define ls now << 1
#define rs now << 1 | 1

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 2e5 + 10;
int n, m;
int val[N], sz[N], dep[N], f[N], son[N], id[N], top[N], a[N], tot;
struct edge{
	int v, nxt;
}e[N << 1];
int head[N], cnt;
void add(int u, int v){
	e[++cnt] = (edge){v, head[u]};
	head[u] = cnt;
}
void dfs1(int u, int fa){
	sz[u] = 1, f[u] = fa, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == fa) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v]) son[u] = v;
	}
}
void dfs2(int u, int t){
	top[u] = t;
	id[u] = ++tot;
	a[tot] = val[u];
	if (son[u]) dfs2(son[u], t);
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == f[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
struct tree{
	int lc, rc, sum, l, r, tag;
}t[N << 2];
void pushup(int now){
	t[now].sum = t[ls].sum + t[rs].sum - (t[ls].rc == t[rs].lc);
	t[now].lc = t[ls].lc, t[now].rc = t[rs].rc;
}
void pushdown(int now){
	if (t[now].tag != -1){
		t[ls].lc = t[ls].rc = t[now].tag;
		t[rs].lc = t[rs].rc = t[now].tag;
		t[ls].sum = t[rs].sum = 1;
		t[ls].tag = t[rs].tag = t[now].tag;
		t[now].tag = -1;
	}
}
void build(int now, int l, int r){
	t[now].l = l, t[now].r = r, t[now].sum = t[now].lc = t[now].rc = 0, t[now].tag = -1;
	if (l == r){
		t[now].lc = t[now].rc = a[l];
		t[now].sum = 1;
		return;
	}int mid = (l + r) >> 1;
	build(ls, l, mid);
	build(rs, mid + 1, r);
	pushup(now);
}
void modify(int now, int x, int y, int k){
	int l = t[now].l, r = t[now].r;
	if (x <= l && r <= y){
		t[now].lc = t[now].rc = t[now].tag = k;
		t[now].sum = 1;
		return;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) modify(ls, x, y, k);
	if (mid + 1 <= y) modify(rs, x, y, k);
	pushup(now);
}
void modify_chain(int x, int y, int k){
	int fx = top[x], fy = top[y];
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		modify(1, id[fx], id[x], k);
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	modify(1, id[x], id[y], k);
}
int query(int now, int x, int y){
	int l = t[now].l, r = t[now].r, res = 0;
	if (x <= l && r <= y){
		return t[now].sum;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) res += query(ls, x, y);
	if (mid + 1 <= y) res += query(rs, x, y);
	if(x <= mid && mid + 1 <= y && t[ls].rc == t[rs].lc) res--;
	return res;
}
int check(int now, int x){
	int l = t[now].l, r = t[now].r;
	if (l == r){
		return t[now].lc;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) return check(ls, x);
	else return check(rs, x);
}
int query_chain(int x, int y){
	int fx = top[x], fy = top[y], res = 0;
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		res += query(1, id[fx], id[x]);
		if (check(1, id[fx]) == check(1, id[f[fx]])) res--;
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	res += query(1, id[x], id[y]);
	return res;
}

int main(){
	n = read(), m = read();
	for (int i = 1; i <= n; i++) val[i] = read();
	for (int i = 1; i < n; i++){
		int u = read(), v = read();
		add(u, v), add(v, u);
	}
	dfs1(1, 0);
	dfs2(1, 1);
	build(1, 1, n);
	for (int i = 1; i <= m; i++){
		char opt;
		cin >> opt;
		int x = read(), y = read(), k;
		if (opt == 'C'){
			k = read();
			modify_chain(x, y, k);
		}else{
			cout << query_chain(x, y) << '\n';
		}
	}
	return 0;
}