Sleeping

Sleeping
시간 제한:
1000ms | 메모리 제한:
65535 KB
난이도:
3
묘사
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.
입력
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.
출력
For each test case, output an integer, indicating the maximal points ZZZ can get.
샘플 입력

10 3 31 2 3 4 5 6 7 8 9 10

샘플 출력

49

제목:

    n  ，ZZZ               ，                L  ，         m       （ m     ），

            ，            。。。

사고방식: 비교적 이해하기 어려운 dp, 특히 tdp[i]가 대표하는 의미.

#include <stdio.h>
#include <string.h>

int dp[1001][1001], tdp[1001], sum[1001];  //dp[i][j]: i   j         ， tdp[j]: i    j      ，        （  i-l i    ）

int mymax(int a, int b)
{
    return a > b ? a : b;
}

int main()
{
    int n, m, l, i, j;
    while(scanf("%d%d%d", &n, &m, &l) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        memset(tdp, 0, sizeof(tdp));
        sum[0] = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &sum[i]);
            sum[i] += sum[i-1];          //     i  
        }
        for(i = 1; i <= n; i++)
        {
            for(j = 0; j <= m && j <= i; j++)
            {
                dp[i][j] = j >= 1 ? dp[i-1][j-1] : 0;    //     i         
                if(i-j >= l)                        //     l   
                {
                     tdp[j] = mymax(tdp[j]+sum[i]-sum[i-1], dp[i-l][j]+sum[i]-sum[i-l]);  //    tdp[j]  （      dp[i-l][j]+...           l    l    ）
                }
                dp[i][j] = mymax(tdp[j], dp[i][j]);   //       
            }
        }
        printf("%d
", dp[n][m]);
    }
    return 0;
}