HDU 3905 Sleeping dp

Sleeping

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1820    Accepted Submission(s): 671

Problem Description

ZZZ is an enthusiastic ACMer and he spends lots of time on training. He always stays up late for training. He needs enough time to sleep, and hates skipping classes. So he always sleeps in the class. With the final exams coming, he has to spare some time to listen to the teacher. Today, he hears that the teacher will have a revision class. The class is N (1 <= N <= 1000) minutes long. If ZZZ listens to the teacher in the i-th minute, he can get Ai points (1<=Ai<=1000). If he starts listening, he will listen to the teacher at least L (1 <= L <= N) minutes consecutively. It`s the most important that he must have at least M (1 <= M <= N) minutes for sleeping (the M minutes needn`t be consecutive). Suppose ZZZ knows the points he can get in every minute. Now help ZZZ to compute the maximal points he can get.

 

Input

The input contains several cases. The first line of each case contains three integers N, M, L mentioned in the description. The second line follows N integers separated by spaces. The i-th integer Ai means there are Ai points in the i-th minute.

 

Output

For each test case, output an integer, indicating the maximal points ZZZ can get.

 

Sample Input

  
  

   
   10 3 3
1 2 3 4 5 6 7 8 9 10
  
  

 

Sample Output

  
  

   
   49
  
  

 

Source

2011 Multi-University Training Contest 7 - Host by ECNU 

璇渣开始补dp Q_Q

n分钟的课 听第i分钟可获得ai分 睡觉不得分

至少睡m分，每次听课必须连续>=L 分钟，求最大得分。

dp[i][j][0] 表示前 i 分钟睡觉了 j 分钟而且第 i 分钟没有睡觉，dp[i][j][1] 表示前 i 分钟睡觉了 j 分钟而且第 i 分钟在睡觉。则：
dp[i][j][1] = max(dp[i-1][j-1][0], dp[i-1][j-1][1]);
dp[i][j][0] = max(dp[i-L][j][1]+sum[i]-sum[i-L], dp[i-1][j][0]+a[i]);

/** Author: ☆·aosaki(*’(OO)’*)  niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 1002

using namespace std;


int dp[1005][1005][2];
int a[1005],sum[1005];

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,L;
    while (~scanf("%d%d%d",&n,&m,&L))
    {
        mem(sum); mem(dp);
        lp(i,n)
          sc(a[i]);
        lp(i,n)
            sum[i]=sum[i-1]+a[i];
        lp(i,n)
        {
            if(i>=L)
                dp[i][0][0]=sum[i];
            for (int j=1;j<=m && j<=i;j++)
            {
                dp[i][j][1]=max(dp[i-1][j-1][0], dp[i-1][j-1][1]);
                if(i-L>=j)
                  dp[i][j][0]=max(dp[i-L][j][1]+sum[i]-sum[i-L], dp[i-1][j][0]+a[i]);
            }
        }
        printf("%d\n",max(dp[n][m][0],dp[n][m][1]));
    }
}