P3572 [POI2014]PTA-Little Bird

題目描述

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!

從1開始，跳到比當前矮的不消耗體力，否則消耗一點體力，每次詢問有一個步伐限制，求每次最少耗費多少體力

輸入輸出格式

輸入格式：

?

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.

?

輸出格式：

?

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.

?

輸入輸出樣例

輸入樣例#1：

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

輸出樣例#1：

2
1

說明

從1開始，跳到比當前矮的不消耗體力，否則消耗一點體力，每次詢問有一個步伐限制，求每次最少耗費多少體力

#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=1;i<=n;i++)scanf("%d",&a[i]);scanf("%d",&q);int limit;for(int i=1;i<=q;i++){scanf("%d",&limit);memset(dp,127/3,sizeof(dp));dp[1]=0;for(int i=2;i<=n;i++){for(int j=i-1;i-j<=limit&&j>=1;j--){if(a[j]>a[i])dp[i]=min(dp[i],dp[j]);else dp[i]=min(dp[i],dp[j]+1);}}printf("%d\n",dp[n]);}
}

40分 暴力dp

/*非常標準的單調隊列優化dp維護兩個單調性，首先是dp[]單調遞減，其次是a[]單調遞增 
*/
#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=1;i<=n;i++)scanf("%d",&a[i]);scanf("%d",&op);int limit;for(int i=1;i<=op;i++){scanf("%d",&limit);h=1,t=1;dp[1]=0;q[1]=1;for(int i=2;i<=n;i++){while(t-h>=0&&i-q[h]>limit)h++;dp[i]=dp[q[h]]+(a[q[h]]<=a[i]);while(t-h>=0&&((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 單調隊列優化dp

?

轉載于:https://www.cnblogs.com/thmyl/p/7494621.html

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