【bzoj3831】[Poi2014]Little Bird 單調隊列優化dp

原文地址：http://www.cnblogs.com/GXZlegend/p/6826475.html

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題目描述

In the Byteotian Line Forest there are   trees in a row. On top of the first one, there is a little bird who would like to fly over to the top of the last tree. Being in fact very little, the bird might lack the strength to fly there without any stop. If the bird is sitting on top of the tree no.  , then in a single flight leg it can fly to any of the trees no.i+1,i+2…I+K, and then has to rest afterward.

Moreover, flying up is far harder to flying down. A flight leg is tiresome if it ends in a tree at least as high as the one where is started. Otherwise the flight leg is not tiresome.

The goal is to select the trees on which the little bird will land so that the overall flight is least tiresome, i.e., it has the minimum number of tiresome legs. We note that birds are social creatures, and our bird has a few bird-friends who would also like to get from the first tree to the last one. The stamina of all the birds varies, so the bird's friends may have different values of the parameter  . Help all the birds, little and big!

有一排n棵樹，第i棵樹的高度是Di。

MHY要從第一棵樹到第n棵樹去找他的妹子玩。

若是MHY在第i棵樹，那麼他能夠跳到第i+1,i+2,...,i+k棵樹。

若是MHY跳到一棵不矮於當前樹的樹，那麼他的勞累值會+1，不然不會。

爲了有體力和妹子玩，MHY要最小化勞累值。

輸入

There is a single integer N(2<=N<=1 000 000) in the first line of the standard input: the number of trees in the Byteotian Line Forest. The second line of input holds   integers D1,D2…Dn(1<=Di<=10^9) separated by single spaces: Di is the height of the i-th tree.

The third line of the input holds a single integer Q(1<=Q<=25): the number of birds whose flights need to be planned. The following Q lines describe these birds: in the i-th of these lines, there is an integer Ki(1<=Ki<=N-1) specifying the i-th bird's stamina. In other words, the maximum number of trees that the i-th bird can pass before it has to rest is Ki-1.

輸出

Your program should print exactly Q lines to the standard output. In the I-th line, it should specify the minimum number of tiresome flight legs of the i-th bird.

樣例輸入

9
4 6 3 6 3 7 2 6 5
2
2
5

樣例輸出

2
1

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題解

單調隊列優化dp

設f[i]表示跳到i的最小勞累值，那麼有f[i]=f[j]+(height[i]>=height[j])

因爲f[i]-f[j]最大爲1，因此從f[k]=f[j]+x轉移過來必定不是最優，即f小的更優。

同時應儘可能讓height[j]大，因此當f相同時，height更大的更優。

據此咱們能夠維護一個單調隊列，其中f單調遞增，f相同時height單調遞減。

而後判斷邊界更新f並加入到隊列中便可。

#include <cstdio>
int a[1000010] , q[1000010] , f[1000010];
bool cmp(int x , int y)
{
	return f[x] == f[y] ? a[x] > a[y] : f[x] < f[y];
}
int main()
{
	int n , i , m , k , l , r;
	scanf("%d" , &n);
	for(i = 1 ; i <= n ; i ++ ) scanf("%d" , &a[i]);
	scanf("%d" , &m);
	while(m -- )
	{
		scanf("%d" , &k);
		l = r = q[1] = 1;
		for(i = 2 ; i <= n ; i ++ )
		{
			while(l <= r && q[l] < i - k) l ++ ;
			f[i] = f[q[l]] + (a[q[l]] <= a[i]);
			while(l <= r && cmp(i , q[r])) r -- ;
			q[++r] = i;
		}
		printf("%d\n" , f[n]);
	}
	return 0;
}