BZOJ 1036 [ZJOI2008]树的统计Count 动态维护树上求和与求最大值 LCT板题

模板,也可以用树链剖分+线段树做O(nlog2)O(nlog^2)O(nlog2)

用LCT做O(nlog)O(nlog)O(nlog)在乘上一个大于30的常数…然后LCT比树剖慢一倍…

CODE

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
template<typename T>inline void read(T &num) {
	char ch; int flg = 1;
	while((ch=getchar())<'0'||ch>'9')if(ch=='-')flg=-flg;
	for(num=0;ch>='0'&&ch<='9';num=num*10+ch-'0',ch=getchar());
	num*=flg;
}
const int MAXN = 30005;
int n, q, u[MAXN], v[MAXN];
namespace LCT {
	#define ls ch[x][0]
	#define rs ch[x][1]
	int ch[MAXN][2], fa[MAXN];
	LL w[MAXN], sum[MAXN], mx[MAXN];
	bool rev[MAXN];
	inline bool isr(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; } //判断是否为根
	inline bool get(int x) { return x == ch[fa[x]][1]; }
	inline void upd(int x) { //上传
		sum[x] = sum[ls] + sum[rs] + w[x];
		mx[x] = max(w[x], max(mx[ls], mx[rs]));
	}
	inline void rot(int x) {
		int y = fa[x], z = fa[y], l = get(x), r = l^1;
		if(!isr(y)) ch[z][get(y)] = x;
		fa[ch[x][r]] = y; fa[y] = x; fa[x] = z;
		ch[y][l] = ch[x][r]; ch[x][r] = y;
		upd(y), upd(x);
	}
	inline void mt(int x) { if(rev[x]) rev[x] ^= 1, rev[ls] ^= 1, rev[rs] ^= 1, swap(ls, rs); } //下传
	void mtpath(int x) { if(!isr(x)) mtpath(fa[x]); mt(x); }
	inline void splay(int x) {
		mtpath(x);
		for(; !isr(x); rot(x))
			if(!isr(fa[x])) rot(get(x)==get(fa[x])?fa[x]:x);
	}
	inline int access(int x) { int y=0;
		for(; x; x=fa[y=x]) splay(x), ch[x][1]=y, upd(x);
		return y;
	}
	inline void bert(int x) { access(x), splay(x), rev[x] ^= 1; } //换根
	inline int sert(int x) { //找根
		access(x), splay(x);
		for(; ch[x][0]; x=ch[x][0]);
		return x;
	}
	inline void link(int x, int y) {
		bert(x);
		if(sert(y) == x) return;
		fa[x] = y;
	}
	inline void cut(int x, int y) {
		bert(x), access(y), splay(y);
		if(sert(y) != x || fa[x] != y || ch[x][1] != 0) return;
		fa[x] = ch[y][0] = 0; upd(y);
	}
	inline void modify(int x, int val) {
		splay(x), w[x] = val, upd(x);
	}
	inline int split(int x, int y) {
		bert(x), access(y), splay(y);
		return y;
	}
	inline int querymax(int x, int y) {
		split(x, y); return mx[y];
	}
	inline int querysum(int x, int y) {
		split(x, y); return sum[y];
	}
}
using namespace LCT;
int main () {
	read(n); mx[0] = -0x7f7f7f7f; //把0设成-无穷,因为upd的时候会访问到
	for(int i = 1; i < n; ++i) read(u[i]), read(v[i]);
	for(int i = 1; i <= n; ++i) read(w[i]);
	for(int i = 1; i < n; ++i) link(u[i], v[i]);
	read(q);
	char s[10]; int x, y;
	while(q--) {
		scanf("%s", s); read(x), read(y);
		if(s[1] == 'M') printf("%d\n", querymax(x, y));
		else if(s[1] == 'S') printf("%d\n", querysum(x, y));
		else modify(x, y);
	}
}