hust 1539 I’m tired

题目描述

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 (1<= Ai <=1000)points. 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 ZZZknows the points he can get in every minute. Now help ZZZ to compute the maximal points he can get.
输入

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 thei-th minute.
输出For each test case, output an integer, indicating the maximal points ZZZ can get.
样例输入

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

样例输出

49

显然，类似于乘积最大，我们想到一个dp方程
dp[i][j]表示在前i分钟睡了j分钟的最优，则dp[i][j]=max(dp[i][j],d[k][j-1]+s[k][i])；
不过很遗憾，这个方程是n^3的时间复杂度，于是我们想优化一下，当然可以优化，具体看程序吧

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int dp[1001][1001],f[1001][1001],sum[1001],a[1001];

int main()
{
    int n,m,L;
    while (scanf("%d%d%d",&n,&m,&L)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(f,0,sizeof(f));
        sum[0]=0;
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
            if (i>=L) dp[i][0]=sum[i];
        }
        for (int i=1;i<=n;i++)
        for (int j=1;j<=min(m,i);j++)
        {
            if (i-j<L) break;
            f[i][j]=max(f[i-1][j]+a[i],dp[i-L][j]+sum[i]-sum[i-L]);
            dp[i][j]=max(dp[i-1][j-1],f[i][j]);
        }
        printf("%d
",dp[n][m]);
    }
    return 0;
}