POJ 3661 Running(dp)

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 2534210

Sample Output

9

Source

USACO 2008 January Silver

dp[i][j]表示第i分钟时 愤怒为 j 所跑的最大距离  
dp[i][j]=dp[i-1][j-1]+d[i]; //第i分钟选择跑
dp[i][0] = dp[i-1][0] //第i分钟选择休息
dp[i][0] = max(dp[i][0],dp[i-k][k])

//从第i-k分钟时疲惫为 k 的状态休息而来 

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define Schar(n) scanf("%c",&n)#define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)using namespace std;typedef long long ll;/*dp[i][j]表示第i分钟时 愤怒为 j 所跑的最大距离  dp[i][j]=dp[i-1][j-1]+d[i]; //第i分钟选择跑dp[i][0] = dp[i-1][0] //第i分钟选择休息dp[i][0] = max(dp[i][0],dp[i-k][k])//从第i-k分钟时疲惫为 k 的状态休息而来 */const int N = 10007;const int M = 507;int dp[N][M]; int d[N];int n,m;int DP(int i,int j){if(i==0&&j==0) return 0;if (i==0) return 0;if (~dp[i][j]) return dp[i][j];if (j == 0){dp[i][0] = DP(i-1,0);for (int k = 1;k <= m;++k){if (i-k>=0) {dp[i][0] = max(dp[i][0],DP(i-k,k));}}return dp[i][j];}  dp[i][j] = DP(i-1,j-1)+d[i];return dp[i][j];}int main(){    while (Sint2(n,m) == 2)    {    for (int i = 1;i <= n;++i) Sint(d[i]);    mem(dp,-1);    Pintc(DP(n,0),'\n');}    return 0;}/*2 101 10*/