（POJ

（POJ - 3661）Running

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 7134 Accepted: 2672

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 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

题目大意：有n分钟，第i分钟可以跑a[i]米，并且增加1个疲劳值，疲劳值不能超过m，当然每一分钟都可以选择休息，休息一分钟可以减少一个疲劳值，一旦休息一定要等到疲劳值为0才能再跑步，要求第n分钟的时候人的疲劳值为0，问这个人能跑的最远距离是多少。

思路：设f[i][j]表示第i分钟疲劳值为j时能跑的最远距离。因为每一分钟都可以选择跑或者休息，则可能状态转移有：①第i分钟选择跑f[i][j]=f[i−1][j−1]+a[i];②在i-1时疲劳值已经为0第i分钟还是选择休息f[i][0]=f[i−1][0];③前面疲劳值为k时的那一刻选择休息一直休息到第i分钟使疲劳值降为0可得f[i][0]=f[i−k][k]（其中i>k)。显然答案就是f[n][0]。

#include<cstdio>#include<algorithm>using namespace std;const int maxn=10005;int a[maxn],f[maxn][505];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        for(int i=1;i<=n;i++) scanf("%d",a+i);        f[0][0]=0;        for(int i=1;i<=n;i++)        {            f[i][0]=f[i-1][0];            for(int j=1;j<=m;j++)                f[i][j]=f[i-1][j-1]+a[i];            for(int k=1;k<=m&&i>k;k++)                f[i][0]=max(f[i][0],f[i-k][k]);        }        printf("%d\n",f[n][0]);    }    return 0;}