POJ 2181 -- Jumping Cows

Posted YXY_1211

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 2181 -- Jumping Cows相关的知识,希望对你有一定的参考价值。

Jumping Cows
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 7624   Accepted: 4586

Description

Farmer John‘s cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump. 

The local witch doctor has mixed up P (1 <= P <= 150,000) potions to aid the cows in their quest to jump. These potions must be administered exactly in the order they were created, though some may be skipped. 

Each potion has a ‘strength‘ (1 <= strength <= 500) that enhances the cows‘ jumping ability. Taking a potion during an odd time step increases the cows‘ jump; taking a potion during an even time step decreases the jump. Before taking any potions the cows‘ jumping ability is, of course, 0. 

No potion can be taken twice, and once the cow has begun taking potions, one potion must be taken during each time step, starting at time 1. One or more potions may be skipped in each turn. 

Determine which potions to take to get the highest jump.

Input

* Line 1: A single integer, P 

* Lines 2..P+1: Each line contains a single integer that is the strength of a potion. Line 2 gives the strength of the first potion; line 3 gives the strength of the second potion; and so on. 

Output

* Line 1: A single integer that is the maximum possible jump. 

Sample Input

8
7
2
1
8
4
3
5
6

Sample Output

17

Source

 
思路:分奇偶考虑,奇数秒时比大,偶数秒时比小。但是目前并没有证明出这就是正确的(虽然找不到更优解)。
技术分享
 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 #define MAXN 150010
 8 using namespace std;
 9 int a[MAXN];
10 int main()
11 {
12     int n;
13     scanf("%d",&n);
14     for(int i=1;i<=n;i++) scanf("%d",&a[i]);
15     int ans=0;
16     bool b=1;
17     for(int i=1;i<=n;i++)
18     {
19         if(b)
20         {
21             if(a[i]>a[i+1]) {ans+=a[i];b=0;}
22             else continue;
23         }
24         else
25         {
26             if(a[i]<a[i+1]) {ans-=a[i];b=1;}
27             else continue;
28         }
29     }
30     printf("%d",ans);
31     return 0;
32 }
POJ 2181

 

以上是关于POJ 2181 -- Jumping Cows的主要内容,如果未能解决你的问题,请参考以下文章

poj 2181 jumping cows

POJ 2181 Jumping Cows DP

百练:2181:Jumping Cows

百练:2181:Jumping Cows

poj2181 jumping cow

HDU 2713: Jumping Cows