【LG3703】[SDOI2017]树点涂色

【LG3703】[SDOI2017]树点涂色

题面

洛谷

题解

一次只能染根到\(x\),且染的颜色未出现过

这句话是我们解题的关键。

\(x\)到根的颜色数为\(f(x)\),则\(u\)\(v\)的颜色数:\(f(u)+f(v)-f(lca_{u,v})+1\)

想一想,为什么?

很显然,如果没有\(1\)操作,我们直接树剖维护一下就可以了。

但是现在有了\(1\)操作。。。

这个\(1\)操作,其实是拉一条从\(x\)到根的链,染成一种颜色

这是不是很像\(LCT\)\(access\)呢?

这样的话,我们就搞一颗\(LCT\)\(access\)时,

因为每断一颗子树,那棵子树内必然要多加一个颜色段就是一个子树加,

每连上一颗子树,那棵子树内必然重复一个颜色段就是一个子树减。

那么我们用树剖维护每个点的\(f(x)\)

并魔改一下\(access\)即可

代码:

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
using namespace std; 
inline int gi() { 
    register int data = 0, w = 1; 
    register char ch = 0; 
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int MAX_N = 1e5 + 5; 
struct Graph { int to, next; } e[MAX_N << 1]; int fir[MAX_N], e_cnt; 
void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; } 
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}, fir[u] = e_cnt++; } 
int N, M; 
int dfn[MAX_N], L[MAX_N], R[MAX_N], top[MAX_N]; 
int dep[MAX_N], fa[MAX_N], son[MAX_N], size[MAX_N], tim; 
void dfs1(int x) { 
	dep[x] = dep[fa[x]] + 1, size[x] = 1; 
	for (int i = fir[x]; ~i; i = e[i].next) { 
		int v = e[i].to; if (v == fa[x]) continue; 
		fa[v] = x; dfs1(v); size[x] += size[v]; 
		if (size[son[x]] < size[v]) son[x] = v; 
	} 
} 
void dfs2(int x, int tp) { 
	top[x] = tp, L[x] = ++tim, dfn[tim] = x; 
	if (son[x]) dfs2(son[x], tp); 
	for (int i = fir[x]; ~i; i = e[i].next) { 
		int v = e[i].to; if (v == fa[x] || v == son[x]) continue; 
		dfs2(v, v); 
	} 
	R[x] = tim; 
}
int LCA(int x, int y) {
	while (top[x] != top[y]) { 
		if (dep[top[x]] < dep[top[y]]) swap(x, y); 
		x = fa[top[x]]; 
	} 
	return dep[x] < dep[y] ? x : y; 
} 
#define lson (o << 1) 
#define rson (o << 1 | 1) 
int val[MAX_N << 2], tag[MAX_N << 2];
void pushup(int o) { val[o] = max(val[lson], val[rson]); } 
void puttag(int o, int v) { tag[o] += v; val[o] += v; } 
void pushdown(int o, int l, int r) {
	if (l == r || !tag[o]) return ; 
	puttag(lson, tag[o]); 
	puttag(rson, tag[o]); 
	tag[o] = 0; 
} 
void build(int o, int l, int r) { 
	if (l == r) return (void)(val[o] = dep[dfn[l]]); 
	int mid = (l + r) >> 1; 
	build(lson, l, mid), build(rson, mid + 1, r); 
	pushup(o); 
} 
void modify(int o, int l, int r, int ql, int qr, int v) {  
	if (ql <= l && r <= qr) return (void)(puttag(o, v)); 
	pushdown(o, l, r); 
	int mid = (l + r) >> 1; 
	if (ql <= mid) modify(lson, l, mid, ql, qr, v);
	if (qr > mid) modify(rson, mid + 1, r, ql, qr, v); 
	pushup(o); 
} 
int query(int o, int l, int r, int ql, int qr) { 
	pushdown(o, l, r); 
	if (ql <= l && r <= qr) return val[o]; 
	int mid = (l + r) >> 1, res = 0; 
	if (ql <= mid) res = max(res, query(lson, l, mid, ql, qr)); 
	if (qr > mid) res = max(res, query(rson, mid + 1, r, ql, qr));
	return res; 
} 

struct Node { int ch[2], fa; bool rev; } t[MAX_N]; 
bool get(int x) { return t[t[x].fa].ch[1] == x; } 
bool nroot(int x) { return t[t[x].fa].ch[0] == x || t[t[x].fa].ch[1] == x; } 
void rotate(int x) { 
	int y = t[x].fa, z = t[y].fa, k = get(x); 
	if (nroot(y)) t[z].ch[get(y)] = x; 
	t[x].fa = z; 
	t[t[x].ch[k ^ 1]].fa = y, t[y].ch[k] = t[x].ch[k ^ 1]; 
	t[y].fa = x, t[x].ch[k ^ 1] = y; 
} 
void splay(int x) { 
	while (nroot(x)) { 
		int y = t[x].fa; 
		if (nroot(y)) get(x) ^ get(y) ? rotate(x) : rotate(y);
		rotate(x); 
	} 
} 
int findroot(int x) { while (t[x].ch[0]) x = t[x].ch[0]; return x; } 
void access(int x) { 
	for (int y = 0; x; y = x, x = t[x].fa) { 
		splay(x); 
		if (t[x].ch[1]) {
			int rt = findroot(t[x].ch[1]);
			modify(1, 1, N, L[rt], R[rt], 1); 
		}
		t[x].ch[1] = y; 
		if (t[x].ch[1]) {
			int rt = findroot(t[x].ch[1]); 
			modify(1, 1, N, L[rt], R[rt], -1); 
		} 
	} 
} 

int main () { 
	clearGraph(); 
	N = gi(), M = gi(); 
	for (int i = 1; i < N; i++) {
		int u = gi(), v = gi(); 
		Add_Edge(u, v); 
		Add_Edge(v, u); 
	} 
	dfs1(1), dfs2(1, 1); 
	for (int x = 2; x <= N; x++) t[x].fa = fa[x]; 
	build(1, 1, N);
	
	while (M--) { 
		int op = gi(); 
		if (op == 1) access(gi()); 
		if (op == 2) {
			int u = gi(), v = gi(), lca = LCA(u, v); 
			printf("%d\n", query(1, 1, N, L[u], L[u]) + query(1, 1, N, L[v], L[v]) 
				- 2 * query(1, 1, N, L[lca], L[lca]) + 1); 
		} 
		if (op == 3) {
			int x = gi(); 
			printf("%d\n", query(1, 1, N, L[x], R[x])); 
		} 
	} 
	return 0; 
} 
posted @ 2019-02-10 19:32  heyujun  阅读(203)  评论(2编辑  收藏  举报