【wikioi】1285 宠物收养所

题目链接：http://www.wikioi.com/problem/1285/

算法：Splay

刚开始看到这题，就注意到特征abs了，并且数据n<=80000显然不能暴力，只能用nlgn的做法，综合起来，似乎只有一个答案：Splay。

将每次插入的点splay到树根，然后找到它的前驱和后继，再取它的绝对值即可，我们记作_abs。那么答案就为sum{_absi}，其中i为领养的人的数量，可以看为宠物的数量。

要注意的：

1. 领养者将会领养特点值为a-b的那只宠物 和 特点值为a-b的那个领养者将成功领养该宠物 告诉我们要取前驱。
2. 同一时间呆在收养所中的，要么全是宠物，要么全是领养者，这些宠物和领养者的个数不会超过10000个。 告诉我们要建两棵Splay树来存剩下的人。
3. 操作人的和操作动物的几乎一样。

==========================14.06.13==========================

原来写的splay的bug太多，已换成数组= = ps：14.07.26又换成指针。。。。>_<

详细看另一篇splay文章，http://www.cnblogs.com/iwtwiioi/p/3537061.html

=================================很久以前==============================

下面放上代码

#include <cstdio>
using namespace std;
#define F(rt) rt-> pa
#define K(rt) rt-> key
#define CH(rt, d) rt-> ch[d]
#define C(rt, d) (K(rt) > d ? 0 : 1)
#define NEW(d) new Splay(d)
#define PRE(rt) F(rt) = CH(rt, 0) = CH(rt, 1) = null

int n, ans, who;

struct Splay {
	Splay* ch[2], *pa;
	int key;
	Splay(int d = 0) : key(d) { ch[0] = ch[1] = pa = NULL; }
};

typedef Splay* tree;
tree null = new Splay, root[2] = {null, null};

void rot(tree& rt, int d) {
	tree k = CH(rt, d^1), u = F(rt); int flag = CH(u, 1) == rt;
	CH(rt, d^1) = CH(k, d); if(CH(k, d) != null) F(CH(k, d)) = rt;
	CH(k, d) = rt; F(rt) = k; rt = k; F(rt) = u;
	if(u != null) CH(u, flag) = k;
}

void splay(tree nod, tree& rt) {
	if(nod == null) return;
	tree pa = F(rt);
	while(F(nod) != pa) {
		if(F(nod) == rt)
			rot(rt, CH(rt, 0) == nod);
		else {
			int d  = CH(F(F(nod)), 0) == F(nod);
			int d2 = CH(F(nod), 0)	  == nod;
			if(d == d2) { rot(F(F(nod)), d); rot(F(nod), d2); }
			else { rot(F(nod), d2); rot(F(nod), d); }
		}
	}
	rt = nod;
}

tree maxmin(tree rt, int d) {
	if(rt == null) return null;
	while(CH(rt, d) != null) rt = CH(rt, d);
	return rt;
}

tree ps(tree rt, int d) {
	if(rt == null) return null;
	rt = CH(rt, d);
	return maxmin(rt, d^1);
}

tree search(tree& rt, int d) {
	if(rt == null) return null;
	tree t = rt;
	while(t != null && K(t) != d) t = CH(t, C(t, d));
	splay(t, rt);
	return t;
}

void insert(tree& rt, int d) {
	tree q = NULL, t = rt;
	while(t != null) q = t, t = CH(t, C(t, d));
	t = new Splay(d);
	PRE(t);
	if(q) F(t) = q, CH(q, C(q, d)) = t;
	else rt = t;
	splay(t, rt);
}

void del(tree& rt) {
	if(rt == null) return;
	tree t = rt;
	if(CH(t, 0) == null) t = CH(rt, 1);
	else {
		t = CH(rt, 0);
		splay(ps(rt, 0), t);
		CH(t, 1) = CH(rt, 1);
		if(CH(rt, 1) != null) F(CH(rt, 1)) = t;
	}
	delete rt;
	F(t) = null;
	rt = t;
}

void init(int key, int d) {
	if(root[d^1] == null) { who = d; insert(root[who], key); return; }
	who = d^1;
	insert(root[who], key);
	tree succ = ps(root[who], 0), pred = ps(root[who], 1);
	int l = 0, r = 0;
	if(succ != null) l = K(root[who]) - K(succ);
	if(pred != null) r = K(pred) - K(root[who]);
	del(root[who]);
	if(succ != null && (pred == null || l <= r)) {
		ans = (ans + l) % 1000000;
		splay(succ, root[who]);
		del(root[who]);
	}
	else if(pred != null && (succ == null || r < l)) {
		ans = (ans + r) % 1000000;
		splay(pred, root[who]);
		del(root[who]);
	}
}

int main() {
	PRE(null);
	scanf("%d", &n);
	int c, b;
	while(~scanf("%d%d", &c, &b)) init(b, c);
	printf("%d", ans);
	return 0;
}