SPOJ QTREE6 lct裸题

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题目链接

岛娘出的题，还是比较容易的

#include <iostream>
#include <fstream>
#include <string>
#include <time.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) {
		putchar('-');
		x = -x;
	}
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
typedef long long ll;
typedef pair<int, int> pii;
const int N = 100005;
const int inf = 10000000;
struct Node *null;
struct Node {
	Node *fa, *ch[2];
	int id;
	int s[2];//s[0] this虚边所连的所有子树的连续白色点数总和（不是链）
	int col;
	int ls[2], rs[2], siz;
	//ls[0]是对于这条链 最上端向下的连续白色点数
	int Ls[2], Rs[2];
	//Ls[0]是对于这棵子树 与最上端点相连的连续白色点数
	bool rev;
	inline void clear(int _col, int _id) {
		fa = ch[0] = ch[1] = null;
		siz = 1;
		rev = 0;
		id = _id;
		col = _col;
		for (int i = 0; i < 2; i++) {
			ls[i] = rs[i] = s[i] = 0;
		}
	}
	inline void push_up() {
		if (this == null)return;
		siz = ch[0]->siz + ch[1]->siz + 1;
		for (int i = 0; i < 2; i++) {
			ls[i] = ch[0]->ls[i], rs[i] = ch[1]->rs[i];
			Ls[i] = ch[0]->Ls[i], Rs[i] = ch[1]->Rs[i];
			if (ch[0]->ls[i] == ch[0]->siz && i == col) {
				ls[i] = ch[0]->siz + 1 + ch[1]->ls[i];
				Ls[i]++;
				Ls[i] += s[i];
				Ls[i] += ch[1]->Ls[i];
			}
			if (ch[1]->rs[i] == ch[1]->siz && i == col) {
				rs[i] = ch[1]->siz + 1 + ch[0]->rs[i];
				Rs[i]++;
				Rs[i] += s[i];
				Rs[i] += ch[0]->Rs[i];
			}
		}
	}
	inline void push_down() {
		if (rev) {
			ch[0]->flip();
			ch[1]->flip();
			rev = 0;
		}
	}
	inline void setc(Node *p, int d) {
		ch[d] = p;
		p->fa = this;
	}
	inline bool d() {
		return fa->ch[1] == this;
	}
	inline bool isroot() {
		return fa == null || fa->ch[0] != this && fa->ch[1] != this;
	}
	inline void flip() {
		if (this == null)return;
		swap(ch[0], ch[1]);
		rev ^= 1;
	}
	inline void go() {//从链头开始更新到this
		if (!isroot())fa->go();
		push_down();
	}
	inline void rot() {
		Node *f = fa, *ff = fa->fa;
		int c = d(), cc = fa->d();
		f->setc(ch[!c], c);
		this->setc(f, !c);
		if (ff->ch[cc] == f)ff->setc(this, cc);
		else this->fa = ff;
		f->push_up();
	}
	inline Node*splay() {
		go();
		while (!isroot()) {
			if (!fa->isroot())
				d() == fa->d() ? fa->rot() : rot();
			rot();
		}
		push_up();
		return this;
	}
	inline Node* access() {//access后this就是到根的一条splay，并且this已经是这个splay的根了
		for (Node *p = this, *q = null; p != null; q = p, p = p->fa) {
			p->splay();
			if (p->ch[1] != null)
				for (int i = 0;i < 2;i++)
				p->s[i] += p->ch[1]->Ls[i];
			if (q != null)
				for (int i = 0; i < 2; i++)
				p->s[i] -= q->Ls[i];
			p->setc(q, 1);
			p->push_up();
		}
		return splay();
	}
	inline Node* find_root() {
		Node *x;
		for (x = access(); x->push_down(), x->ch[0] != null; x = x->ch[0]);
		return x;
	}
	void make_root() {
		access()->flip();
	}
	void cut() {//把这个点的子树脱离出去
		access();
		ch[0]->fa = null;
		ch[0] = null;
		push_up();
	}
	void cut(Node *x) {
		if (this == x || find_root() != x->find_root())return;
		else {
			x->make_root();
			cut();
		}
	}
	void link(Node *x) {
		if (find_root() == x->find_root())return;
		else {
			make_root(); fa = x;
		}
	}
};
Node pool[N], *tail;
Node *node[N];
int n, q;
struct Edge {
	int to, nex;
}edge[N << 1];
int head[N], edgenum;
void add(int u, int v) {
	Edge E = { v, head[u] };
	edge[edgenum] = E;
	head[u] = edgenum++;
}
void dfs(int u, int fa) {
	for (int i = head[u]; ~i; i = edge[i].nex) {
		int v = edge[i].to; if (v == fa)continue;
		node[v]->fa = node[u];
		dfs(v, u);
		for (int j = 0; j < 2; j++)
		node[u]->s[j] += node[v]->Ls[j];
	}
	node[u]->push_up();
}
int main() {
	while (cin >> n) {
		memset(head, -1, sizeof head); edgenum = 0;
		for (int i = 1, u, v; i < n; i++) {
			rd(u); rd(v); 
			add(u, v);add(v, u);
		}
		tail = pool; 
		null = tail++;
		null->clear(-1, 0); null->siz = 0;
		for (int i = 1; i <= n; i++)
		{
			node[i] = tail++;
			node[i]->clear(1, i);
		}
		dfs(1, 1);
		rd(q); int u, v;
		while (q--) {
			rd(u); rd(v);
			if (!u) {
				node[v]->access();
				pt(max(node[v]->Rs[0], node[v]->Rs[1])); puts("");
			}
			else {
				node[v]->access();
				node[v]->col ^= 1;
				node[v]->push_up();
			}
		}
	}
	return 0;
}

SPOJ QTREE6 lct裸题

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原文地址：http://blog.csdn.net/qq574857122/article/details/46862921