jzoj 6308. 中间值

Description

详见OJ

Solution

考场先想到(O(nlog^2n))的线段树，发现过不了。于是开始“异想天开”。
最后神奇想到分块。
我们对于(a[])维护一个(to[i])。
(to[i])表示(a[i]>=b[j])的最大的(j)。
维护时用分块来标记防止修改的区间太大。
赛后同学说分块是(O(m根号n))的，我才发现时间好像过不了。。。
但我好像没有一个点(TLE)。。。
不停改细节最后成功(AC)，(700+ms)没有卡线。
分块打法好！

Code

（分块）

#include <cstdio>
#include <cmath>
#define N 500010
#define mem(x, a) memset(x, a, sizeof x)
#define fo(x, a, b) for (int x = a; x <= b; x++)
#define fd(x, a, b) for (int x = a; x >= b; x--)
using namespace std;
int n, m, a[N], b[N], to[N], now = 0;
int bl[N], val[N], st, zuo[N], you[N];
bool bz[N];

inline int read()
{
	int x = 0; char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
	return x;
}

inline int max(int x, int y) {return x > y ? x : y;}

inline int solve(int x) {return bz[bl[x]] ? val[bl[x]] : to[x];}

void gave(int x)
{
	if (! bz[x]) return;
	fo(i, zuo[x], you[x]) to[i] = val[x];
	bz[x] = 0;
}

int main()
{
	freopen("median.in", "r", stdin);
	freopen("median.out", "w", stdout);
	n = read(), m = read(); st = sqrt(n);
	fo(i, 1, n)
	{
		a[i] = read();
		bl[i] = (i - 1) / st + 1;
		if (! zuo[bl[i]]) zuo[bl[i]] = i;
		you[bl[i]] = i;
	}
	fo(i, 1, n) b[i] = read();
	fo(i, 1, n)
	{
		while (now < n && a[i] >= b[now + 1]) now++;
		to[i] = now;
	}
	to[0] = 0, to[n + 1] = n;
	while (m--)
	{
		int opt = read();
		if (opt == 1)
		{
			int x = read(), y = read(), z = read();
			if (x == 0)
			{
				int l = solve(y - 1), r = solve(y + 1), mid;
				while (l <= r)
				{
					mid = l + r >> 1;
					if (z >= b[mid]) l = mid + 1;
					else r = mid - 1;
				}
				a[y] = z;
				if (solve(y) != l - 1)
					to[y] = l - 1, bz[bl[y]] = 0;
			}
			else
			{
				int l = 1, r = n, mid, le, ri;
				while (l <= r)
				{
					mid = l + r >> 1;
					if (a[mid] >= b[y - 1]) r = mid - 1;
					else l = mid + 1;
				}
				le = r + 1;
				l = r, r = n;
				while (l <= r)
				{
					mid = l + r >> 1;
					if (a[mid] < b[y + 1]) l = mid + 1;
					else r = mid - 1;
				}
				ri = l - 1;
				l = le, r = ri;
				while (l <= r)
				{
					mid = l + r >> 1;
					if (a[mid] >= z) r = mid - 1;
					else l = mid + 1;
				}
				if (le <= r)
				{
					if (bl[le] == bl[r])
					{
						gave(bl[le]);
						fo(i, le, r) to[i] = y - 1;
					}
					else
					{
						fo(i, bl[le] + 1, bl[r] - 1)
							bz[i] = 1, val[i] = y - 1;
						gave(bl[le]);
						fo(i, le, you[bl[le]]) to[i] = y - 1;
						gave(bl[r]);
						fo(i, zuo[bl[r]], r) to[i] = y - 1;
					}
				}
				r++;
				if (r <= ri)
				{
					if (bl[r] == bl[ri])
					{
						gave(bl[r]);
						fo(i, r, ri) to[i] = y;
					}
					else
					{
						fo(i, bl[r] + 1, bl[ri] - 1)
							bz[i] = 1, val[i] = y;
						gave(bl[r]);
						fo(i, r, you[bl[r]]) to[i] = y;
						gave(bl[ri]);
						fo(i, zuo[bl[ri]], ri) to[i] = y;
					}
				}
				b[y] = z;
			}
		}
		else
		{
			int l1 = read(), r1 = read(), l2 = read(), r2 = read();
			int sum = (r1 - l1 + 1 + r2 - l2 + 1) / 2 + 1;
			if (a[r1] <= b[l2])
			{
				if (r1 - l1 + 1 >= sum) printf("%d
", a[l1 + sum - 1]);
				else printf("%d
", b[l2 + sum - (r1 - l1 + 1) - 1]);
			}
			else if (a[l1] >= b[r2])
			{
				if (r2 - l2 + 1 >= sum) printf("%d
", b[l2 + sum - 1]);
				else printf("%d
", a[l1 + sum - (r2 - l2 + 1) - 1]);
			}
			else
			{
				int l = l1, r = r1, mid, num;
				while (l <= r)
				{
					mid = l + r >> 1;
					num = mid - l1 + 1;
					if (solve(mid) > r2) num += r2 - l2 + 1;
					else if (solve(mid) >= l2) num += solve(mid) - l2 + 1;
					if (num == sum) {r = mid - 1; break;}
					else if (num >= sum) r = mid - 1;
					else l = mid + 1;
				}
				r++;
				num = r - l1 + 1;
				if (solve(r) > r2) num += r2 - l2 + 1;
				else if (solve(r) >= l2) num += solve(r) - l2 + 1;
				if (r <= r1 && num == sum) printf("%d
", a[r]);
				else printf("%d
", b[l2 + sum - (r - l1 + 1)]);
			}
		}
	}
	return 0;
}

不过，听完讲后，发现这题分治做起来好容易啊。