poj3661 Running

Description

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 D_i (1 ≤ D_i ≤ 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: D_i

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

这题用dp[i][j]表示前i秒疲劳值为j所能走的最远距离。注意一点：dp[i][j](j!=0)只能从dp[i-1][j-1]+a[i]转移过来，不能从dp[i-1][j+1]转移过来，因为如果上一秒的疲劳度大于1，那么这一秒的疲劳度必不为0，即这个转移过来的值不能继续走，而是要等到它为0时再走。                                                                                                                                                                            

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
int a[10060],dp[10060][505];
int main()
{
    int n,m,i,j,maxx;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        memset(dp,-1,sizeof(dp));
        dp[1][0]=0;
        dp[1][1]=a[1];
        for(i=2;i<=n;i++){
            dp[i][0]=max(dp[i-1][1],dp[i-1][0]);
            for(j=1;j<=m;j++){
                if(dp[i-1][j-1]!=-1){
                    dp[i][j]=max(dp[i][j],dp[i-1][j-1]+a[i]);
                }
                if(i>j &&dp[i-j][j]!=-1){
                    dp[i][0]=max(dp[i-j][j],dp[i][0]);
                }
            }
        }
        printf("%d
",dp[n][0]);
    }
    return 0;
}