HDU 5002 Tree

题意：

一棵树  支持删边加边、路径权值加值、路径权值改值、路径求第二大的数字和其个数

思路：

LCT的第二题  题意已经把功能都告诉了  比較裸

要注意的是权值加值和改值两个操作的标记下放问题  要先down改值  再down加值

对于路径的操作通过mroot变换树的形态再access拿出路径比較方便  不要像我上一篇一样搞lca

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<cstdlib>
#include<ctime>
#include<cmath>
using namespace std;
#define N 100010
#define L(x) (ch[x][0])
#define R(x) (ch[x][1])
#define inf -2147483640

int ch[N][2], pre[N], rev[N], me[N];
int cov[N], add[N], size[N], max1[N], num1[N], max2[N], num2[N];
bool rt[N];

struct helper {
	int val, num;
	bool operator<(const helper ff) const {
		return val > ff.val;
	}
} hel[10];

void Update_COV(int u, int d) {
	if (!u)
		return;
	me[u] = d;
	cov[u] = d;
	add[u] = inf;
	max1[u] = d;
	num1[u] = size[u];
	max2[u] = inf;
	num2[u] = 0;
}

void Update_Add(int u, int d) {
	if (!u)
		return;
	me[u] += d;
	if (add[u] != inf)
		add[u] += d;
	else
		add[u] = d;
	max1[u] += d;
	if (max2[u] != inf)
		max2[u] += d;
}

void Update_Rev(int u) {
	if (!u)
		return;
	swap(L(u), R(u));
	rev[u] ^= 1;
}

void down(int u) {
	if (rev[u]) {
		Update_Rev(L(u));
		Update_Rev(R(u));
		rev[u] = 0;
	}
	if (cov[u] != inf) {
		Update_COV(L(u), cov[u]);
		Update_COV(R(u), cov[u]);
		cov[u] = inf;
	}
	if (add[u] != inf) {
		Update_Add(L(u), add[u]);
		Update_Add(R(u), add[u]);
		add[u] = inf;
	}
}

void up(int u) {
	size[u] = size[L(u)] + size[R(u)] + 1;
	hel[0].val = max1[L(u)];
	hel[0].num = num1[L(u)];
	hel[1].val = max2[L(u)];
	hel[1].num = num2[L(u)];
	hel[2].val = max1[R(u)];
	hel[2].num = num1[R(u)];
	hel[3].val = max2[R(u)];
	hel[3].num = num2[R(u)];
	hel[4].val = me[u];
	hel[4].num = 1;
	sort(hel, hel + 5);
	int i;
	for (i = 1; i < 5; i++) {
		if (hel[i].val != hel[i - 1].val)
			break;
	}
	max1[u] = hel[0].val;
	max2[u] = hel[i].val;
	num1[u] = num2[u] = 0;
	for (i = 0; i < 5; i++) {
		if (hel[i].val == max1[u])
			num1[u] += hel[i].num;
		else if (hel[i].val == max2[u])
			num2[u] += hel[i].num;
	}
}

//Rotate P Splay 一般不变
void Rotate(int x) {
	int y = pre[x], kind = ch[y][1] == x;
	ch[y][kind] = ch[x][!kind];
	pre[ch[y][kind]] = y;
	pre[x] = pre[y];
	pre[y] = x;
	ch[x][!kind] = y;
	if (rt[y])
		rt[y] = false, rt[x] = true;
	else
		ch[pre[x]][ch[pre[x]][1] == y] = x;
	up(y);
}

//P函数先将splay根结点到u的路径上全部的结点的标记逐级下放
void P(int u) {
	if (!rt[u])
		P(pre[u]);
	down(u);
}

void Splay(int u) {
	P(u);
	while (!rt[u]) {
		int fa = pre[u], ffa = pre[fa];
		if (rt[fa])
			Rotate(u);
		else if ((R(ffa) == fa) == (R(fa) == u))
			Rotate(fa), Rotate(u);
		else
			Rotate(u), Rotate(u);
	}
	up(u);
}

//将root到u的路径变成实边
int Access(int u) {
	int v = 0;
	for (; u; u = pre[v = u]) {
		Splay(u);
		rt[R(u)] = true, rt[R(u) = v] = false;
		up(u);
	}
	return v;
}

//使u成为它所在的树的根
void mroot(int u) {
	Access(u);
	Splay(u);
	Update_Rev(u);
}

//连接两棵树  u接在v上
void link(int u, int v) {
	mroot(u);
	pre[u] = v;
}

//u-v边断开
void cut(int u, int v) {
	mroot(u);
	Access(u);
	Splay(v);
	pre[L(v)] = pre[v];
	pre[v] = 0;
	rt[L(v)] = true;
	L(v) = 0;
	up(v);
}

//u-v路径权值变为w
void COV(int u, int v, int w) {
	mroot(u);
	Access(v);
	Splay(v);
	Update_COV(v, w);
}

//u-v路径+w
void ADD(int u, int v, int w) {
	mroot(u);
	Access(v);
	Splay(v);
	Update_Add(v, w);
}

//u-v路径最大值
void query(int u, int v) {
	mroot(u);
	Access(v);
	Splay(v);
	if (max2[v] != inf)
		printf("%d %d
", max2[v], num2[v]);
	else
		printf("ALL SAME
");
}

struct edge {
	int v, next;
} ed[N * 2];
int head[N], tot;

void addedge(int u, int v) {
	ed[tot].v = v;
	ed[tot].next = head[u];
	head[u] = tot++;
}

//利用dfs初始化pre数组  建立LCT
void dfs(int u) {
	for (int i = head[u]; ~i; i = ed[i].next) {
		int v = ed[i].v;
		if (pre[v] != 0)
			continue;
		pre[v] = u;
		dfs(v);
	}
}

int main() {
	int T, t, n, m, i, op, u, v, x, y, w;
	max1[0] = max2[0] = inf;
	rt[0] = true;
	scanf("%d", &T);
	for (t = 1; t <= T; t++) {
		scanf("%d%d", &n, &m);
		for (i = 1; i <= n; i++) {
			scanf("%d", &me[i]);
			max1[i] = me[i];
			max2[i] = inf;
			num1[i] = 1;
			num2[i] = 0;
			size[i] = 1;
			add[i] = cov[i] = inf;
			L(i) = R(i) = pre[i] = rev[i] = 0;
			rt[i] = true;
		}
		tot = 0;
		memset(head, -1, sizeof(head));
		for (i = 1; i < n; i++) {
			scanf("%d%d", &u, &v);
			addedge(u, v);
			addedge(v, u);
		}
		pre[1] = -1;
		dfs(1);
		pre[1] = 0;
		printf("Case #%d:
", t);
		while (m--) {
			scanf("%d", &op);
			if (op == 1) {
				scanf("%d%d%d%d", &u, &v, &x, &y);
				cut(u, v);
				link(x, y);
			} else if (op == 2) {
				scanf("%d%d%d", &u, &v, &w);
				COV(u, v, w);
			} else if (op == 3) {
				scanf("%d%d%d", &u, &v, &w);
				ADD(u, v, w);
			} else {
				scanf("%d%d", &u, &v);
				query(u, v);
			}
		}
	}
	return 0;
}