本文主要是介绍POJ - 3661，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Running
The cows are trying to become better athletes, so Bessie is running on a track for exactly N (1 ≤ N ≤ 10,000) minutes. During each minute, she can choose to either run or rest for the whole minute.
The ultimate distance Bessie runs, though, depends on her ‘exhaustion factor’, which starts at 0. When she chooses to run in minute i, she will run exactly a distance of Di (1 ≤ Di ≤ 1,000) and her exhaustion factor will increase by 1 – but must never be allowed to exceed M (1 ≤ M ≤ 500). If she chooses to rest, her exhaustion factor will decrease by 1 for each minute she rests. She cannot commence running again until her exhaustion factor reaches 0. At that point, she can choose to run or rest.
At the end of the N minute workout, Bessie’s exaustion factor must be exactly 0, or she will not have enough energy left for the rest of the day.
Find the maximal distance Bessie can run.
Input
*Line 1: Two space-separated integers: N and M
*Lines 2…N+1: Line i+1 contains the single integer: Di
Output
*Line 1: A single integer representing the largest distance Bessie can run while satisfying the conditions.
Sample Input
5 2
5
3
4
2
10
Sample Output
9
经典dp题 题意跑步给出当前分钟跑步时会前进的距离 每分钟跑步的时候会增长一点疲劳点疲劳点不能超过一定的限度(每次都可以选择是否休息)。
#include <iostream>
#include <cmath>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
int a[100009],dp[10009][509],c,n;
int inf=1e9;
int main()
{int k,u,m;while(cin>>n>>c){memset(dp,0,sizeof(dp));for(int i=1; i<=n; i++){cin>>a[i];}for(int i=1; i<=n; i++)for(int j=0;j<=c&&j<=i;j++){if(j==0)dp[i][0]=dp[i-1][0];else{dp[i][j]=max(dp[i-1][j-1]+a[i],dp[i][j]);dp[i][0]=max(dp[i][0],dp[i-j][j]);}}cout<<dp[n][0]<<endl;}
}
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