「HNOI2004」宠物收养场

「HNOI2004」宠物收养场

传送门

对宠物和人各维护一棵平衡树，每次 \(\text{split}\) 的时候记得判一下子树是否为空，然后模拟就好了。

参考代码：

#include <algorithm>
#include <cstdlib>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T & s) {
	s = 0; int f = 0; char c = getchar();
	while ('0' > c || c > '9') f |= c == '-', c = getchar();
	while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
	s = f ? -s : s;
}

const int _ = 8e4 + 5, mod = 1e6;

int n, rt1, rt2, tot, siz[_], val[_], pri[_], ch[2][_];

inline int Newnode(int v)
{ return siz[++tot] = 1, val[tot] = v, pri[tot] = rand(), tot; }

inline void pushup(int p) { siz[p] = siz[ch[0][p]] + siz[ch[1][p]] + 1; }

inline int merge(int x, int y) {
	if (!x || !y) return x + y;
	if (pri[x] > pri[y])
		return ch[1][x] = merge(ch[1][x], y), pushup(x), x;
	else
		return ch[0][y] = merge(x, ch[0][y]), pushup(y), y;
}

inline void split(int p, int v, int& x, int& y) {
	if (!p) { x = y = 0; return ; }
	if (val[p] <= v)
		return x = p, split(ch[1][p], v, ch[1][x], y), pushup(p);
	else
		return y = p, split(ch[0][p], v, x, ch[0][y]), pushup(p);
}

inline int kth(int p, int k) {
	if (siz[ch[0][p]] + 1 > k) return kth(ch[0][p], k);
	if (siz[ch[0][p]] + 1 == k) return val[p];
	if (siz[ch[0][p]] + 1 < k) return kth(ch[1][p], k - siz[ch[0][p]] - 1);
}

inline void insert(int& rt, int item) {
	int a, b;
	split(rt, item, a, b);
	rt = merge(a, merge(Newnode(item), b));
}

inline void erase(int& rt, int item) {
	int a, b, c;
	split(rt, item, a, c);
	split(a, item - 1, a, b);
	b = merge(ch[0][b], ch[1][b]);
	rt = merge(a, merge(b, c));
}

inline int match(int& rt, int item) {
	int a, b, x, y, res;
	split(rt, item, a, b);
	if (a == 0 && b != 0) res = kth(b, 1);
	if (a != 0 && b == 0) res = kth(a, siz[a]);
	if (a != 0 && b != 0)
		x = kth(a, siz[a]), y = kth(b, 1), res = item - x <= y - item ? x : y;
	rt = merge(a, b);
	return res;
}

int main() {
#ifndef ONLINE_JUDGE
	file("cpp");
#endif
	read(n);
	int ans = 0;
	for (rg int a, b, x, i = 1; i <= n; ++i) {
		read(a), read(b);
		if (a == 0) {
			if (rt2 == 0) insert(rt1, b);
			else x = match(rt2, b), erase(rt2, x), ans = (ans + abs(x - b)) % mod;
		} else {
			if (rt1 == 0) insert(rt2, b);
			else x = match(rt1, b), erase(rt1, x), ans = (ans + abs(x - b)) % mod;
		}
	}
	printf("%d\n", ans);
	return 0;
}