P2486 [SDOI2011]染色

(color{#0066ff}{ 题目描述 })

(color{#0066ff}{输入格式})

(color{#0066ff}{输出格式})

对于每个询问操作，输出一行答案。

(color{#0066ff}{输入样例})

6 5
2 2 1 2 1 1
1 2
1 3
2 4
2 5
2 6
Q 3 5
C 2 1 1
Q 3 5
C 5 1 2
Q 3 5

(color{#0066ff}{输出样例})

3
1
2

(color{#0066ff}{数据范围与提示})

(color{#0066ff}{ 题解 })

树剖

线段树维护：区间两端点的颜色，区间颜色段数

区间合并的时候判断一下端点颜色即可

树剖的时候一起要判断端点处，那变量记录一下重链的上下，就是线段树区间query[l,r]的两端点颜色

如果相同要--

#include<bits/stdc++.h>
#define LL long long
LL in() {
	char ch; LL x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
const int maxn = 1e5 + 10;
struct node {
	int to;
	node *nxt;
	node(int to = 0, node *nxt = NULL): to(to), nxt(nxt) {}
	void *operator new(size_t) {
		static node *S = NULL, *T = NULL;
		return (S == T) && (T = (S = new node[1024]) + 1024), S++;
	}
};
struct Tree {
protected:
	int lc, rc;
	struct node {
		node *ch[2];
		int l, r, num, tag;
		int lc, rc;
		node(int l = 0, int r = 0, int num = 1, int tag = -1, int lc = 0, int rc = 0)
			:l(l), r(r), num(num), tag(tag), lc(lc), rc(rc) { ch[0] = ch[1] = NULL; }
		void upd() {
			num = ch[0]->num + ch[1]->num;
			if(ch[0]->rc == ch[1]->lc) num--;
			lc = ch[0]->lc;
			rc = ch[1]->rc;
		}
		void trn(int c) {
			num = 1;
			lc = rc = tag = c;
		}
		void dwn() {
			if(~tag) {
				ch[0]->trn(tag);
				ch[1]->trn(tag);
				tag = -1;
			}
		}
	};
	node *root;
	void build(node *&o, int l, int r, int *val, int *rev) {
		o = new Tree::node(l, r, 1, -1, 0, 0);
		if(l == r) return (void)(*o = Tree::node(l, r, 1, -1, val[rev[l]], val[rev[l]]));
		int mid = (l + r) >> 1;
		build(o->ch[0], l, mid, val, rev);
		build(o->ch[1], mid + 1, r, val, rev);
		o->upd();
	}
	void lazy(node *o, int l, int r, int c) {
		if(o->r < l || o->l > r) return;
		if(l <= o->l && o->r <= r) {
			o->tag = o->lc = o->rc = c;
			o->num = 1;
			return;
		}
		o->dwn();
		lazy(o->ch[0], l, r, c), lazy(o->ch[1], l, r, c);
		o->upd();
	}
	int query(node *o, int l, int r) {
		if(o->r < l || o->l > r) return 0;
		if(o->l == l) lc = o->lc;
		if(o->r == r) rc = o->rc;
		if(l <= o->l && o->r <= r) return o->num;
		o->dwn();
		int lans = query(o->ch[0], l, r);
		int rans = query(o->ch[1], l, r);
		o->upd();
		if(lans && rans && o->ch[0]->rc == o->ch[1]->lc) return lans + rans - 1;
		return lans + rans;
	}
public:
	int L() { return lc; }
	int R() { return rc; }
	void build(int l, int r, int *val, int *rev) { build(root, l, r, val, rev); }
	void lazy(int l, int r, int c) { lazy(root, l, r, c); }
	int query(int l, int r) { return query(root, l, r); }
}t;
node *head[maxn];
int top[maxn], dfn[maxn], redfn[maxn], son[maxn];
int siz[maxn], fa[maxn], val[maxn], cnt, dep[maxn];
int n, m;
void add(int from, int to) { head[from] = new node(to, head[from]); }
char getch() {
	char ch;
	while(!isalpha(ch = getchar()));
	return ch;
}
void dfs1(int x, int f) {
	fa[x] = f;
	dep[x] = dep[f] + 1;
	siz[x] = 1;
	for(node *i = head[x]; i; i = i->nxt) {
		if(i->to == f) continue;
		dfs1(i->to, x);
		siz[x] += siz[i->to];
		if(!son[x] || siz[i->to] > siz[son[x]]) son[x] = i->to;
	}
}
void dfs2(int x, int t) {
	top[redfn[dfn[x] = ++cnt] = x] = t;
	if(son[x]) dfs2(son[x], t);
	for(node *i = head[x]; i; i = i->nxt)
		if(!dfn[i->to]) 
			dfs2(i->to, i->to);
}
void addpath(int x, int y, int c) {
	int fx = top[x], fy = top[y];
	while(fx != fy) {
		if(dep[fx] >= dep[fy]) {
			t.lazy(dfn[fx], dfn[x], c);
			x = fa[fx];
		}
		else {
			t.lazy(dfn[fy], dfn[y], c);
			y = fa[fy];
		}
		fx = top[x];
		fy = top[y];
	}
	if(dep[x] > dep[y]) t.lazy(dfn[y], dfn[x], c);
	else t.lazy(dfn[x], dfn[y], c);
}
int querypath(int x, int y) {
	int ans = 0, ansl = -1, ansr = -1;
	int fx = top[x], fy = top[y];
	while(fx != fy) {
		if(dep[fx] >= dep[fy]) {
			ans += t.query(dfn[fx], dfn[x]);
			if(t.R() == ansl) ans--;
			ansl = t.L();
			x = fa[fx];
		}
		else {
			ans += t.query(dfn[fy], dfn[y]);
			if(t.R() == ansr) ans--;
			ansr = t.L();
			y = fa[fy];
		}
		fx = top[x];
		fy = top[y];
	}
	if(dep[x] > dep[y]) {
		ans += t.query(dfn[y], dfn[x]);
		if(t.L() == ansr) ans--;
		if(t.R() == ansl) ans--;
	}
	else {
		ans += t.query(dfn[x], dfn[y]);
		if(t.L() == ansl) ans--;
		if(t.R() == ansr) ans--;
	}
	return ans;
}
int main() {
	n = in(), m = in();
	for(int i = 1; i <= n; i++) val[i] = in();
	int x, y, z;
	for(int i = 1; i < n; i++) x = in(), y = in(), add(x, y), add(y, x);
	dfs1(1, 0), dfs2(1, 1), t.build(1, n, val, redfn);
	while(m --> 0) {
		if(getch() == 'C') {
			x = in(), y = in(), z = in();
			addpath(x, y, z);
		}
		else {
			x = in(), y = in();
			printf("%d
", querypath(x, y));
		}
	}
	return 0;
}