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树上染色 / 染色

题目链接：ybt金牌导航5-1-3 / luogu P2486

题目大意

给你一个树，然后点有颜色，要你维护两个操作。
把一条路径上的点的颜色改成某个值，查询一条路径的颜色段数量。
颜色段是指最长的颜色相同的段。

思路

看到树上路径，自然想到树链剖分。

然后你发现你就需要线段树维护区间最左最右的颜色，这个很好维护。
然后你也可以维护出区间颜色段的。
然后记得树链剖分的时候你将链合并时它们也可能缝合，那你每次要把这条链两端颜色找出来，然后拿去比较。

然后大概就是这样。

代码

#include<cstdio>
#include<algorithm>

using namespace std;

struct node {
	int to, nxt;
}e[200001];
int n, m, a[100001], x, y, z;
int le[100001], KK, lcol, rcol;
int fa[100001], son[100001], sz[100001], deg[100001];
int pl[100001], dfn[100001], top[100001], tmp;
char op;
struct Tree {
	int lc, rc, num, lazy;
}tr[400001];

void add(int x, int y) {
	e[++KK] = (node){y, le[x]}; le[x] = KK;
	e[++KK] = (node){x, le[y]}; le[y] = KK;
}

void dfs1(int now, int father) {//树链剖分
	fa[now] = father;
	deg[now] = deg[father] + 1;
	sz[now] = 1;
	int maxn = 0;
	for (int i = le[now]; i; i = e[i].nxt)
		if (e[i].to != father) {
			dfs1(e[i].to, now);
			sz[now] += sz[e[i].to];
			if (sz[e[i].to] > maxn) {
				maxn = sz[e[i].to];
				son[now] = e[i].to;
			}
		}
}

void dfs2(int now, int father) {
	if (son[now]) {
		pl[son[now]] = ++tmp;
		top[son[now]] = top[now];
		dfn[tmp] = son[now];
		dfs2(son[now], now);
	}
	
	for (int i = le[now]; i; i = e[i].nxt)
		if (e[i].to != father && e[i].to != son[now]) {
			pl[e[i].to] = ++tmp;
			top[e[i].to] = e[i].to;
			dfn[tmp] = e[i].to;
			dfs2(e[i].to, now);
		}
}

//线段树操作
void up(int now) {
	tr[now].lc = tr[now << 1].lc;
	tr[now].rc = tr[now << 1 | 1].rc;
	tr[now].num = tr[now << 1].num + tr[now << 1 | 1].num;
	if (tr[now << 1].rc == tr[now << 1 | 1].lc)//接合的地方下相同颜色，两颜色段合并，所以个数少了一个
		tr[now].num--;//下面的很多地方减一都是处理这个
}

void down(int now, int l, int r) {
	if (tr[now].lazy != -1 && l != r) {
		tr[now << 1].num = tr[now << 1 | 1].num = 1;
		tr[now << 1].lc = tr[now << 1].rc = tr[now].lazy;
		tr[now << 1 | 1].lc = tr[now << 1 | 1].rc = tr[now].lazy;
		tr[now << 1].lazy = tr[now << 1 | 1].lazy = tr[now].lazy;
		tr[now].lazy = -1;
	}
}

void build(int now, int l, int r) {
	if (l == r) {
		tr[now] = (Tree){a[dfn[l]], a[dfn[l]], 1, -1};
		return ;
	}
	
	tr[now].lazy = -1;
	
	int mid = (l + r) >> 1;
	build(now << 1, l, mid);
	build(now << 1 | 1, mid + 1, r);
	
	up(now);
}

void change(int now, int l, int r, int L, int R, int num) {
	if (L <= l && r <= R) {
		tr[now].lc = tr[now].rc = num;
		tr[now].num = 1;
		tr[now].lazy = num;
		return ;
	}
	
	down(now, l, r);
	int mid = (l + r) >> 1;
	if (L <= mid) change(now << 1, l, mid, L, R, num);
	if (mid < R) change(now << 1 | 1, mid + 1, r, L, R, num);
	
	up(now);
}

int query(int now, int l, int r, int L, int R, int &l_col, int &r_col) {
	if (L <= l && r <= R) {
		if (l == L) l_col = tr[now].lc;//找到这段区间最左最右的颜色
		if (r == R) r_col = tr[now].rc;
		return tr[now].num;
	}
	
	down(now, l, r);
	int mid = (l + r) >> 1, re = 0;
	if (L <= mid) re += query(now << 1, l, mid, L, R, l_col, r_col);
	if (mid < R) re += query(now << 1 | 1, mid + 1, r, L, R, l_col, r_col);
	if (L <= mid && mid < R) {
		if (tr[now << 1].rc == tr[now << 1 | 1].lc) re--;//相同颜色缝合
	}
	
	return re;
}

void color(int x, int y, int z) {
	int X = top[x], Y = top[y];
	while (X != Y) {
		if (deg[X] < deg[Y]) {
			swap(X, Y);
			swap(x, y); 
		}
		change(1, 1, tmp, pl[X], pl[x], z);
		x = fa[X];
		X = top[x];
	}
	
	if (deg[x] > deg[y]) swap(x, y);
	change(1, 1, tmp, pl[x], pl[y], z);
}

int ask_num(int x, int y) {
	int X = top[x], Y = top[y], re = 0, tmp1, tmp2;
	lcol = rcol = -2;
	while (X != Y) {
		if (deg[X] < deg[Y]) {
			swap(X, Y);
			swap(x, y);
			swap(lcol, rcol);
		}
		tmp1 = tmp2 = -4;
		re += query(1, 1, tmp, pl[X], pl[x], tmp1, tmp2);
		if (tmp2 == lcol) re--;//跟你前面走的链缝合
		lcol = tmp1;
		x = fa[X];
		X = top[x];
	}
	
	if (deg[x] > deg[y]) swap(x, y), swap(lcol, rcol);
	tmp1 = tmp2 = -4;
	re += query(1, 1, tmp, pl[x], pl[y], tmp1, tmp2);
	if (tmp1 == lcol) re--;//对接处有两个可以缝合的地方
	if (tmp2 == rcol) re--;
	
	return re;
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	
	for (int i = 1; i < n; i++) {
		scanf("%d %d", &x, &y);
		add(x, y);
	}
	
	dfs1(1, 0);
	pl[1] = ++tmp; top[1] = 1; dfn[tmp] = 1;
	dfs2(1, 0);
	
	build(1, 1, n);
	
	while (m--) {
		op = getchar();
		while (op != ‘C‘ && op != ‘Q‘) op = getchar();
		
		if (op == ‘C‘) {
			scanf("%d %d %d", &x, &y, &z);
			color(x, y, z);
			continue;
		}
		if (op == ‘Q‘) {
			scanf("%d %d", &x, &y);
			printf("%d\n", ask_num(x, y));
			continue;
		}
	}
	
	return 0;
}

【ybt金牌导航5-1-3】【luogu P2486】树上染色 / 染色

标签：查询   node   return   amp   std   --   etc   ==   scan   

原文地址：https://www.cnblogs.com/Sakura-TJH/p/YBT_JPDH_5-1-3.html