bzojP3512: [POI2010]PIL-Pilots

Pre

一道不太难的蓝题，有一点需要注意。

Solution

四次单调队列(少几次也可以，但是需要空间支持)。

Code

#include<bits/stdc++.h>
#define xx first
#define yy second
#define ll long long
using namespace std;

const int N = 3000000 + 5;
ll info[N], n, q[N], tail, head;
ll l[N], r[N], k;

int main () {
	memset (r, 127, sizeof (r));
	scanf ("%lld%lld", &k, &n);
	for (int i = 1; i <= n; ++i) {
		scanf ("%lld", &info[i]);
	}
	
	l[1] = 1;
	tail = head = 1;
	q[1] = 1;
	for (int i = 2; i <= n; ++i) {
		l[i] = l[i - 1];
		while (tail >= head && info[q[head]] - info[i] > k) {
			l[i] = max (q[head] + 1, l[i]);
			head++;
		}
		while (tail >= head && info[q[tail]] < info[i]) {
			tail--;
		}
		q[++tail] = i;
	}
	
	tail = head = 1;
	q[1] = 1;
	for (int i = 2; i <= n; ++i) {
		l[i] = max (l[i], l[i - 1]);
		while (tail >= head && info[q[head]] - info[i] < -k) {
			l[i] = max (l[i], q[head] + 1);
			head++;
		}
		while (tail >= head && info[q[tail]] >= info[i]) {
			tail--;
		}
		q[++tail] = i;
	}
	
	tail = head = 1;
	q[1] = n;
	r[n] = n;
	for (int i = n - 1; i >= 1; --i) {
		r[i] = r[i + 1];
		while (tail >= head && info[q[head]] - info[i] > k) {
			r[i] = min (q[head] - 1, r[i]);
			head++;
		}
		while (tail >= head && info[q[tail]] <= info[i]) {
			tail--;
		}
		q[++tail] = i;
	}
	
	tail = head = 1;
	q[1] = n;
	for (int i = n - 1; i >= 1; --i) {
		r[i] = min (r[i], r[i + 1]);
		while (tail >= head && info[q[head]] - info[i] < -k) {
			r[i] = min (r[i], q[head] - 1);
			head++;
		}
		while (tail >= head && info[q[tail]] >= info[i]) {
			tail--;
		}
		q[++tail] = i;
	}
	ll ans = 1, L = 1;
	for (int i = 2; i <= n; ++i) {
		L = max (L, l[i]);
		ans = max (ans, i - L + 1);
	}
	printf ("%lld
", ans);
	return 0;
}

Conclusion

第一次敲出代码样例都没有过就提交了，直接10分(竟然还有10分)。

发现有一个地方，在算出了限制位置(就是对于每一个点在区间内时，所选取的区间的最远的端点)之后，不能够直接

for (int i = 1; i <= n; ++i) {
    ans = max (ans, r[i] - l[i] + 1);
}

因为区间内的数是非常有可能排斥的，正确的做法是

for (int i = 1; i <= n; ++i) {
	L = max (L, l[i]);
	ans = max (ans, i - L + 1);
}

就是枚举每一个点作为右端点，(DP)出左端点的最远取值。