洛谷 P2486 染色

因为多加了一个等号，然后100->0，开个博客纪念一下

题目大意

自己看去，我就不信你题目都读不懂

solution

求的是连续的颜色段的个数.
因为给出的111221的颜色段为三.
我们考虑怎么从下边上传.
如果我们是从111 221 来的，因为中间没有相同的可以接加上左右儿子的颜色段.
但是如果是这样11 12 那么合并的时候需要判断一下左右儿子相连 的那一部分是不是相同.
如果相同那么需要-1，不相同则不用-1.然后我们就能写出push_up 啦.

push_up

obviously

void push_up(int rt) {
	tree[rt].lc = tree[lson].lc;
	tree[rt].rc = tree[rson].rc;
	tree[rt].sum = tree[lson].sum + tree[rson].sum;
	if (tree[lson].rc == tree[rson].lc) tree[rt].sum--;
}

因为区间查询的时候也需要上传，那么我们就需要判断所求区间是不是需要合并一下子，
如果需要合并那么还要判断是不是需要-1的那种情况.
那么我们query也有了.

query

obviously

void query(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) {
		ans += tree[rt].sum;
		if (l == L) LC = tree[rt].lc;
		if (r == R) RC = tree[rt].rc;
		return;
	}
	push_down(rt);
	int mid = (l + r) >> 1;
	if (R <= mid) query(lson, l, mid, L, R);
	else if (L > mid) query(rson, mid + 1, r, L, R);
	else {
		query(lson, l, mid, L, R);
		query(rson, mid + 1, r, L, R);
		if (tree[lson].rc == tree[rson].lc) ans--;
	}
}

还有树剖的时候向上跳，显然我们需要维护上一个区间的lc和rc来判断当前需不需要减一

int ask(int x, int y) {
	int res = 0, x1 = 0, y1 = 0; 
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) swap(x, y), swap(x1, y1);
		ans = 0;
		Seg::query(1, 1, n, dfn[top[x]], dfn[x]);
		res += ans;
		if (RC == x1) res--;
		x1 = LC, x = fath[top[x]];
	}
	if (dfn[x] < dfn[y]) swap(x, y), swap(x1, y1);
	ans = 0;
	Seg::query(1, 1, n, dfn[y], dfn[x]);
	res += ans;	
	if (LC == y1) res--;
	if (RC == x1) res--;
	return res;
}

code

/*
	Auther:_Destiny
	time:2020.5.9
*/
#include <bits/stdc++.h>
#define N 100010
#define M 1010

using namespace std;
struct TTT {
	int lc, rc, sum, lazy;
}tree[N << 2];
int n, m, ans, LC, RC;
int dfn[N], dep[N], fath[N], w[N], pre[N], son[N], top[N], siz[N];

int read() {
	int s = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

namespace Seg {
	#define lson rt << 1
	#define rson rt << 1 | 1
	void push_up(int rt) {
		tree[rt].lc = tree[lson].lc;
		tree[rt].rc = tree[rson].rc;
		tree[rt].sum = tree[lson].sum + tree[rson].sum;
		if (tree[lson].rc == tree[rson].lc) tree[rt].sum--;
	}
	void build(int rt, int l, int r) {
		if (l == r) {
			tree[rt].lc = tree[rt].rc = w[pre[l]];
			tree[rt].sum = 1;
			return;
		}
		int mid = (l + r) >> 1;
		build(lson, l, mid);
		build(rson, mid + 1, r);
		push_up(rt);
	}
	void push_down(int rt) {
		if (!tree[rt].lazy) return;
		tree[lson].lazy = tree[rson].lazy = tree[rt].lazy;
		tree[lson].sum = tree[rson].sum = 1;
		tree[lson].lc = tree[lson].rc = tree[rt].lazy;
		tree[rson].lc = tree[rson].rc = tree[rt].lazy;
		tree[rt].lazy = 0;
	}
	void update(int rt, int c, int l, int r, int L, int R) {
		if (L <= l && r <= R) {
			tree[rt].lazy = c;
			tree[rt].lc = tree[rt].rc = c;
			tree[rt].sum = 1;
			return;
		}
		push_down(rt);
		int mid = (l + r) >> 1;
		if (L <= mid) update(lson, c, l, mid, L, R);
		if (R > mid) update(rson, c, mid + 1, r, L, R);
		push_up(rt);
	}
	void query(int rt, int l, int r, int L, int R) {
		if (L <= l && r <= R) {
			ans += tree[rt].sum;
			if (l == L) LC = tree[rt].lc;
			if (r == R) RC = tree[rt].rc;
			return;
		}
		push_down(rt);
		int mid = (l + r) >> 1;
		if (R <= mid) query(lson, l, mid, L, R);
		else if (L > mid) query(rson, mid + 1, r, L, R);
		else {
			query(lson, l, mid, L, R);
			query(rson, mid + 1, r, L, R);
			if (tree[lson].rc == tree[rson].lc) ans--;
		}
	}
}

namespace Cut {
	int add_edge, cnt, head[N << 1];
	struct node {
		int next, to;
	}edge[N << 1];
	void add(int from, int to) {
		edge[++add_edge].next = head[from];
		edge[add_edge].to = to;
		head[from] = add_edge;
	}
	void dfs(int x, int fa) {
		siz[x] = 1, fath[x] = fa, dep[x] = dep[fa] + 1;
		for (int i = head[x]; i; i = edge[i].next) {
			int to = edge[i].to;
			if (to == fa) continue;
			dfs(to, x); siz[x] += siz[to];
			if (siz[son[x]] < siz[to]) son[x] = to;
		}
	}
	void dfs2(int x, int tp) {
		dfn[x] = ++cnt, pre[cnt] = x, top[x] = tp;
		if (son[x]) dfs2(son[x], tp);
		for (int i = head[x]; i; i = edge[i].next) {
			int to = edge[i].to;
			if (to == son[x] || to == fath[x]) continue;
			dfs2(to, to);
		}
	}
	void change(int x, int y, int d) {
		while (top[x] != top[y]) {
			if (dep[top[x]] < dep[top[y]]) swap(x, y);
			Seg::update(1, d, 1, n, dfn[top[x]], dfn[x]);
			x = fath[top[x]];
		}
		if (dfn[x] > dfn[y]) swap(x, y);
		Seg::update(1, d, 1, n, dfn[x], dfn[y]);
	}
	int ask(int x, int y) {
		int res = 0, x1 = 0, y1 = 0; 
		while (top[x] != top[y]) {
			if (dep[top[x]] < dep[top[y]]) swap(x, y), swap(x1, y1);
			ans = 0;
			Seg::query(1, 1, n, dfn[top[x]], dfn[x]);
			res += ans;
			if (RC == x1) res--;
			x1 = LC, x = fath[top[x]];
		}
		if (dfn[x] < dfn[y]) swap(x, y), swap(x1, y1);
		ans = 0;
		Seg::query(1, 1, n, dfn[y], dfn[x]);
		res += ans;	
		if (LC == y1) res--;
		if (RC == x1) res--;
		return res;
	}
}

int main() {
	char s;
	n = read(), m = read();
	for (int i = 1; i <= n; i++) w[i] = read();
	for (int i = 1, x, y; i < n; i++) {
		x = read(), y = read();
		Cut::add(x, y), Cut::add(y, x);
	}
	Cut::dfs(1, 1); Cut::dfs2(1, 1);
	Seg::build(1, 1, n);
	for (int i = 1, x, y, d; i <= m; i++) {
		cin >> s;
		x = read(), y = read();
		if (s == 'C') {
			d = read();
			Cut::change(x, y, d);
		}
		if (s == 'Q') printf("%d\n", Cut::ask(x, y));
	}
}