洛谷 P2486 [SDOI2011]染色

树链剖分裸题。

线段树统计“答案”、“前缀字符”、“后缀字符”三个信息就可以很方便的合并了。

代码：

#include<bits/stdc++.h>

using namespace std;

const int maxn = 100000 + 5;

int n, m;
int head[maxn], nxt[maxn * 4], to[maxn * 4], cnt;
void inline addEdge(int u, int v) {
	nxt[++cnt] = head[u], to[cnt] = v, head[u] = cnt;
	nxt[++cnt] = head[v], to[cnt] = u, head[v] = cnt;
}

namespace Tree
{
	// PartI. Tree
	int col[maxn], son[maxn], fa[maxn], top[maxn], size[maxn], deep[maxn], dfn[maxn], org[maxn], dfs_clock;
	void dfs1(int u, int pa) {
		fa[u] = pa; deep[u] = deep[pa] + 1;
		size[u] = 1;
		for(int e = head[u]; e; e = nxt[e]) {
			int v = to[e];
			if(v == pa) continue;
			dfs1(v, u);
			size[u] += size[v];
			if(size[v] > size[son[u]]) son[u] = v;
		}
	}
	void dfs2(int u, int pa) {
		dfn[u] = ++dfs_clock;
		org[dfs_clock] = u;
		if(son[u]) {
			top[son[u]] = top[u];
			dfs2(son[u], u);
		}
		for(int e = head[u]; e; e = nxt[e]) {
			int v = to[e];
			if(v == son[u] || v == pa) continue;
			top[v] = v;
			dfs2(v, u);
		}
	}
	// PartII. Segment
	struct Node {
		int val;
		int color[2];
		Node() { val = color[0] = color[1] = 0; }
	} node[maxn * 4];
	int tag[maxn * 4];
	Node inline merge(Node left, Node right)
	{
		Node ans;
		ans.val = left.val + right.val - (left.color[1] == right.color[0]);
		ans.color[0] = left.color[0], ans.color[1] = right.color[1];
		return ans;
	}
	void inline pushup(int id) {
		node[id] = merge(node[id << 1], node[id << 1 | 1]);
	}
	void inline pushdown(int id) {
		if(tag[id]) {
			tag[id << 1] = tag[id << 1 | 1] = tag[id];
			node[id << 1].val = node[id << 1 | 1].val = 1;
			node[id << 1].color[0] = node[id << 1].color[1] = node[id << 1 | 1].color[0] = node[id << 1 | 1].color[1] = tag[id];
			tag[id] = 0;
		}
	}
	void build(int id, int l, int r) {
		if(l == r) {
			node[id].val = 1, node[id].color[0] = node[id].color[1] = col[org[l]];
			return;
		}
		int mid = l + r >> 1;
		build(id << 1, l, mid);
		build(id << 1 | 1, mid + 1, r);
		pushup(id);
	}
	Node query(int id, int l, int r, int s, int t) {
		if(s <= l && t >= r) {
			return node[id];
		}
		int mid = l + r >> 1;
		pushdown(id); Node ansl, ansr;
		if(s <= mid) { ansl = query(id << 1, l, mid, s, t); if(t <= mid) return ansl; }
		if(t > mid) { ansr = query(id << 1 | 1, mid + 1, r, s, t); if(s > mid) return ansr; }
		return merge(ansl, ansr);
	}
	void rset(int id, int l, int r, int s, int t, int v) {
		if(s <= l && t >= r) {
			tag[id] = node[id].color[0] = node[id].color[1] = v;
			node[id].val = 1;
			return;
		}
		int mid = l + r >> 1;
		pushdown(id);
		if(s <= mid) rset(id << 1, l, mid, s, t, v);
		if(t > mid) rset(id << 1 | 1, mid + 1, r, s ,t ,v);
		pushup(id);
	}
}

using namespace Tree;

void inline Init()
{
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; ++i) scanf("%d", col + i);
	int u, v;
	for(int i = 1; i < n; ++i) {
		scanf("%d %d", &u, &v);
		addEdge(u, v);
	}
	dfs1(1, 0);
	top[1] = 1;
	dfs2(1, 0);
	build(1, 1, n);
}

char opt[5]; int x, y, v;

void inline Solve()
{
	while(m--) {
		scanf("%s %d %d", opt, &x, &y);
		if(opt[0] == 'C') {
			scanf("%d", &v);
			while(top[x] != top[y]) {
				if(deep[top[x]] < deep[top[y]]) swap(x, y);
				rset(1, 1, n, dfn[top[x]], dfn[x], v);
				x = fa[top[x]];
			}
			if(deep[x] > deep[y]) swap(x, y);
			rset(1, 1, n, dfn[x], dfn[y], v);
		} else {
			Node ansl, ansr;
			while(top[x] != top[y]) {
				if(deep[top[x]] > deep[top[y]]) {
					ansl = merge(query(1, 1, n, dfn[top[x]], dfn[x]), ansl);
					x = fa[top[x]];
				} else {
					ansr = merge(query(1, 1, n, dfn[top[y]], dfn[y]), ansr);
					y = fa[top[y]];
				}
			}
			if(deep[x] > deep[y]) {
				ansl = merge(query(1, 1, n, dfn[y], dfn[x]), ansl);
			} else {
				ansr = merge(query(1 ,1, n, dfn[x], dfn[y]), ansr);
			}
			swap(ansl.color[0], ansl.color[1]);
			printf("%d\n", merge(ansl, ansr).val);
		}
	}
}

int main()
{
	Init();
	Solve();
	return 0;
}