洛谷P3572 [POI2014]PTA-Little Bird

P3572 [POI2014]PTA-Little Bird

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!

从1开始，跳到比当前矮的不消耗体力，否则消耗一点体力，每次询问有一个步伐限制，求每次最少耗费多少体力

输入输出格式

输入格式：

There is a single integer nn (2\le n\le 1\ 000\ 0002≤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 nn integers d_1,d_2,\cdots,d_nd​1​​,d​2​​,⋯,d​n​​ (1\le d_i\le 10^91≤d​i​​≤10​9​​)separated by single spaces: d_id​i​​ is the height of the i-th tree.

The third line of the input holds a single integer qq (1\le q\le 251≤q≤25): the number of birds whoseflights need to be planned.

The following qq lines describe these birds: in the ii-th of these lines, there is an integer k_ik​i​​ (1\le k_i\le n-11≤k​i​​≤n−1) specifying the ii-th bird's stamina. In other words, the maximum number of trees that the ii-th bird can pass before it has to rest is k_i-1k​i​​−1.

输出格式：

Your program should print exactly qq lines to the standard output.

In the ii-th line, it should specify the minimum number of tiresome flight legs of the ii-th bird.

输入输出样例

输入样例#1：

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

输出样例#1：

2
1

说明

从1开始，跳到比当前矮的不消耗体力，否则消耗一点体力，每次询问有一个步伐限制，求每次最少耗费多少体力

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 1000010
using namespace std;
int n,a[maxn],dp[maxn],q;
int main(){
    scanf("%d",&n);
    for(int i=;i<=n;i++)scanf("%d",&a[i]);
    scanf("%d",&q);
    int limit;
    for(int i=;i<=q;i++){
        scanf("%d",&limit);
        memset(dp,/,sizeof(dp));
        dp[]=;
        for(int i=;i<=n;i++){
            for(int j=i-;i-j<=limit&&j>=;j--){
                if(a[j]>a[i])dp[i]=min(dp[i],dp[j]);
                else dp[i]=min(dp[i],dp[j]+);
            }
        }
        printf("%d\n",dp[n]);
    }
}

40分 暴力dp

/*
    非常标准的单调队列优化dp
    维护两个单调性，首先是dp[]单调递减，其次是a[]单调递增
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 1000010
using namespace std;
int n,op,a[maxn],dp[maxn],q[maxn],h,t;
int main(){
    scanf("%d",&n);
    for(int i=;i<=n;i++)scanf("%d",&a[i]);
    scanf("%d",&op);
    int limit;
    for(int i=;i<=op;i++){
        scanf("%d",&limit);
        h=,t=;
        dp[]=;q[]=;
        for(int i=;i<=n;i++){
            while(t-h>=&&i-q[h]>limit)h++;
            dp[i]=dp[q[h]]+(a[q[h]]<=a[i]);
            while(t-h>=&&((dp[q[t]]>dp[i])||(dp[q[t]]==dp[i]&&a[q[t]]<a[i])))t--;
            q[++t]=i;
        }
        printf("%d\n",dp[n]);
    }
}

100分 单调队列优化dp