【POI2014】【BZOJ3831】Little Bird

在Byteotian Line Forest中,一只小鸟需要从第一棵树飞到最后一棵树,通过优化飞行路径来最小化飞行的劳累值。每个飞行段都会根据高度变化累加劳累值,而休息则发生在飞行结束后。此外,小鸟还有朋友需要一起完成这段旅程,因此需要为所有小鸟制定最优化的飞行计划。

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Description

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棵树,第i棵树的高度是Di。
MHY要从第一棵树到第n棵树去找他的妹子玩。
如果MHY在第i棵树,那么他可以跳到第i+1,i+2,…,i+k棵树。
如果MHY跳到一棵不矮于当前树的树,那么他的劳累值会+1,否则不会。
为了有体力和妹子玩,MHY要最小化劳累值。
Input

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.
Output

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.
Sample Input

9

4 6 3 6 3 7 2 6 5

2

2

5
Sample Output

2

1

HINT

Explanation: The first bird may stop at the trees no. 1, 3, 5, 7, 8, 9. Its tiresome flight legs will be the one from the 3-rd tree to the 5-th one and from the 7-th to the 8-th.

Source

鸣谢zhonghaoxi

这题挺sb的..
简单的dp,用单调队列维护一下..
但是我WA了2遍(摔

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 1000010
#define GET (ch>='0'&&ch<='9')
using namespace std;
void in(int &x)
{
    char ch=getchar();x=0;
    while (!GET)    ch=getchar();
    while (GET) x=x*10+ch-'0',ch=getchar();
}
int n,Q,k,i;
int a[MAXN],f[MAXN];
int que[MAXN],head,tail;
int main()
{
    for (in(n),i=1;i<=n;i++)    in(a[i]);
    for (in(Q);Q;Q--)
    {
        in(k);que[head=tail=1]=1;
        for (i=2;i<=n;i++)
        {
            while   (head<=tail&&i-que[head]>k) head++;
            f[i]=f[que[head]]+(a[i]>=a[que[head]]);
            while   (head<=tail&&(f[i]<f[que[tail]]||(f[i]==f[que[tail]]&&a[i]>=a[que[tail]]))) tail--;
            que[++tail]=i;
        }
        printf("%d\n",f[n]);
    }
}
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