_bzoj1208 [HNOI2004]宠物收养所【Splay】

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1208

以后在空间限制允许的情况下我绝对不纠结内存占用问题啦！就因为不舍得用long long而用unsigned，爆掉了我好几个小时。

#include <cstdio>

const int maxn = 80005, delta[2] = {-1, 1}, mod = 1000000;

int n, t1, t2, s, ans;
int ch[maxn][2], fa[maxn], root, cnt = 2;
long long key[maxn], smaller, larger;

inline void rotate(int x) {
	int y = fa[x];
	if (y == ch[fa[y]][0]) {
		ch[fa[y]][0] = x;
	}
	else {
		ch[fa[y]][1] = x;
	}
	fa[x] = fa[y];
	int dir = x == ch[y][1];
	ch[y][dir] = ch[x][dir ^ 1];
	fa[ch[x][dir ^ 1]] = y;
	ch[x][dir ^ 1] = y;
	fa[y] = x;
}
inline void splay(int x) {
	int p;
	while (fa[x]) {
		p = fa[x];
		if (!fa[p]) {
			rotate(x);
		}
		else {
			if ((p == ch[fa[p]][1]) ^ (x == ch[p][1])) {
				rotate(x);
			}
			else {
				rotate(p);
			}
			rotate(x);
		}
	}
	root = x;
}
inline void ist(int x, long long val) {
	int p = 0, dir = 0;
	while (x) {
		p = x;
		dir = val > key[x];
		x = ch[x][dir];
	}
	++cnt;
	ch[p][dir] = cnt;
	fa[cnt] = p;
	key[cnt] = val;
	splay(cnt);
}
inline void del(long long val) {
	int x = root;
	while (key[x] != val) {
		if (val < key[x]) {
			x = ch[x][0];
		}
		else {
			x = ch[x][1];
		}
	}
	splay(x); 
	//注意：此题splay完这个x后，x绝对有两个孩子。
	fa[ch[x][0]] = fa[ch[x][1]] = 0;
	int i;
	for (i = ch[x][0]; ch[i][1]; i = ch[i][1]); 
	splay(i);
	ch[i][1] = ch[x][1];
	fa[ch[x][1]] = i;
}
inline long long dayudengyu(long long val) {
	int x = root;
	long long rt = 0;
	while (x && key[x] != val) {
		if (val < key[x]) {
			rt = key[x];
			x = ch[x][0];
		}
		else {
			x = ch[x][1];
		}
	}
	return x? val: rt;
}
inline long long xiaoyu(long long val) {
	int x = root;
	long long rt = 0;
	while (x) {
		if (val <= key[x]) {
			x = ch[x][0];
		}
		else {
			rt = key[x];
			x = ch[x][1];
		}
	}
	return rt;
}

int main(void) {
	//freopen("in.txt", "r", stdin);
	scanf("%d", &n);
	key[1] = -999999999999999LL;
	ch[1][1] = 2;
	key[2] = 999999999999999LL;
	fa[2] = 1;
	root = 1;
	while (n--) {
		scanf("%d%d", &t1, &t2);
		if (!s) {
			ist(root, t2);
		}
		else {
			if (t1 ^ (s < 0)) {
				ist(root, t2);
			}
			else {
				larger = dayudengyu(t2);
				smaller = xiaoyu(t2);
				if (larger - (long long)t2 < (long long)t2 - smaller) {
					ans = (ans + ((larger - (long long)t2) % mod)) % mod;
					del(larger);
				}
				else {
					ans = (ans + (((long long)t2 - smaller) % mod)) % mod;
					del(smaller);
				}
			}
		}
		s += delta[t1];
	}
	printf("%d\n", ans);
	return 0;
}