[POI2014]PTA-Little Bird 解题报告

传送门：https://www.luogu.org/problemnew/show/P3572

题目描述

In the Byteotian Line Forest there are nn 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. i, then in a single flight leg it can fly toany of the trees no. i+1,i+2,\cdots,i+ki+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 leastas 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 leasttiresome, 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 getfrom 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 kk .

Help all the birds, little and big!

输入输出格式

输入格式：

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 n 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 whoseflights 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 .

输出格式：

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.

输入输出样例

输入样例#1： 复制

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

输出样例#1： 复制

2
1

题目大意是：只鸟要飞过一片含棵树的森林，第只鸟每飞一步最多只能飞过棵树；给出每棵树的高度，若从第棵树飞到第棵树且，则需要消耗体力，否则不需要消耗体力。问每只鸟飞过树林消耗的最少体力。

朴素的想法是用表示飞到第i棵树消耗的最少体力，可由……转移而来，即：

这样子枚举，时间复杂度为，怎么都超时。

所以必须用到单调队列优化dp。

如果，或者且，显然比优。

好了，那么就可以根据这个构造单调队列了，时间复杂度为。

代码：

#include
#include
using namespace std;
    int f[1000001];
    int n,q;
    int a[1000001];
    int que[1000001];
    int head,tail;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    scanf("%d",&q);
    for (int i=1;i<=q;i++)
    {
        int k;
        scanf("%d",&k);
        head=1;
        tail=1;
        que[head]=1;
        int r=2;
        memset(f,0,sizeof(f));
        while (r<=n)
        {
            while (headf[que[head+1]]||(f[que[head]]==f[que[head+1]]&&a[que[head]]a[r]) f[r]=f[que[head]]; else f[r]=f[que[head]]+1; //状态转移
            while (head<=tail&&(f[r]a[que[tail]]))) tail--; //更新队尾
            que[++tail]=r;
            r++;
        }
        printf("%d\n",f[n]);
    }
}