洛谷 P2486 [SDOI2011]染色（树链剖分+线段树）

题目链接

题解

比较裸的树链剖分 好像树链剖分的题都很裸

线段树中维护一个区间最左和最右的颜色，和答案

合并判断一下中间一段就可以了

比较考验代码能力

Code

#include<bits/stdc++.h>
#define LL long long
#define RG register
using namespace std;

inline int gi() {
    int f = 1, s = 0;
    char c = getchar();
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    if (c == '-') f = -1, c = getchar();
    while (c >= '0' && c <= '9') s = s*10+c-'0', c = getchar();
    return f == 1 ? s : -s;
}
const int N = 100010;
struct node {
	int to, next;
}g[N<<1];
int last[N], gl;
inline void add(int x, int y) {
	g[++gl] = (node) {y, last[x]};
	last[x] = gl;
	g[++gl] = (node) {x, last[y]};
	last[y] = gl;
	return ;
}

int c[N];

struct Segment_tree {
	int l, r, v, lazy;
}t[N<<2];

int fa[N], siz[N], son[N], cnt, s[N], top[N], id[N], dep[N];
void dfs1(int u, int f) {
	siz[u] = 1; fa[u] = f;
	int MAX = 0;
	for (int i = last[u]; i; i = g[i].next) {
		int v = g[i].to;
		if (v == f) continue;
		dep[v] = dep[u]+1;
		dfs1(v, u);
		siz[u] += siz[v];
		if (MAX < siz[v]) MAX = siz[v], son[u] = v;
	}
	return ;
}

void dfs2(int u, int topf) {
	top[u] = topf;
	s[++cnt] = c[u];
	id[u] = cnt;
	if (!son[u]) return ;
	dfs2(son[u], topf);
	for (int i = last[u]; i; i = g[i].next) {
		int v = g[i].to;
		if (v == fa[u] || v == son[u]) continue;
			dfs2(v, v);
	}
	return ;
}
#define mid ((l+r)>>1)
#define ls (rt<<1)
#define rs (rt<<1|1)

void pushup(int rt) {
	t[rt].l = t[ls].l; t[rt].r = t[rs].r;
	t[rt].v = t[ls].v + t[rs].v;
	if (t[ls].r == t[rs].l) t[rt].v--;
	return ;
}

void build(int rt, int l, int r) {
	if (l == r) {
		t[rt] = (Segment_tree) {s[l], s[l], 1, 0};
		return ;
	}
	build(ls, l, mid); build(rs, mid+1, r);
	pushup(rt);
	return ;
}
void pushdown(int rt) {
	int lazy = t[rt].lazy;
	t[rt].lazy = 0;
	if (lazy) {
		t[rs].lazy = t[ls].lazy = lazy;
		t[rs].l = t[rs].r = t[ls].l = t[ls].r = lazy;
		t[rs].v = t[ls].v = 1;
	}
	return ;
}
void update(int rt, int l, int r, int L, int R, int k) {
	if (L <= l && r <= R) {
		t[rt].v = 1;
		t[rt].lazy = t[rt].l = t[rt].r = k;
		return ;
	}
	pushdown(rt);
	if (L <= mid) update(ls, l, mid, L, R, k);
	if (R > mid) update(rs, mid+1, r, L, R, k);
	pushup(rt);
	return ;
}

struct zzy {
	int l, r, v;
}; 

zzy merge(zzy a, zzy b) {
	if (!a.v) return b;
	if (!b.v) return a;
	return (zzy) {a.l, b.r, a.v+b.v-(a.r == b.l)};
}

zzy query(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R)
		return (zzy) {t[rt].l, t[rt].r, t[rt].v};
	zzy z = (zzy) {0, 0, 0};
	pushdown(rt);
	if (L <= mid)
		z = query(ls, l, mid, L, R);
	if (R > mid)
		z = merge(z, query(rs, mid+1, r, L, R));
	return z;
}

void upway(int x, int y, int z) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		update(1, 1, cnt, id[top[x]], id[x], z);
		x = fa[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	update(1, 1, cnt, id[x], id[y], z);
	return ;
}

int qway(int x, int y) {
	zzy zx, zy;
	zx = zy = (zzy) {0, 0, 0};
	while (top[x] != top[y]) {
		if (dep[top[x]] >= dep[top[y]]) {
			zx = merge(query(1, 1, cnt, id[top[x]], id[x]), zx);
			x = fa[top[x]];
		}
		else {
			zy = merge(query(1, 1, cnt, id[top[y]], id[y]), zy);
			y = fa[top[y]];
		}
	}
	if (dep[x] > dep[y]) swap(x, y), swap(zx, zy);
	zy = merge(query(1, 1, cnt, id[x], id[y]), zy);
	zx.v += zy.v;
	if (zx.l == zy.l) zx.v--;
	return zx.v;
}

int main() {
	int n = gi(), m = gi();
	for (int i = 1; i <= n; i++) c[i] = gi();
	for (int i = 1; i < n; i++) add(gi(), gi());
	dfs1(1, 0);
	dfs2(1, 1);
	build(1, 1, n);
	while (m--) {
		char c; cin>>c;
		if (c == 'C') {
			int x = gi(), y = gi(), z = gi();
			upway(x, y, z);
		}
		else {
			int x = gi(), y = gi();
			printf("%d\n", qway(x, y));
		}
	}
    return 0;
}