题目描述

The cows are trying to become better athletes, so Bessie is running on a track for exactly 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 and her exhaustion factor will increase by 1 -- but must never be allowed to exceed . 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.

输入描述:

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 contains the single integer:

输出描述:

* Line 1: A single integer representing the largest distance Bessie can run while satisfying the conditions.

示例1

输入
5 2
5
3
4
2
10
输出
9
说明
Bessie runs during the first minute (distance=5), rests during the second minute, runs for the third (distance=4), and rests for the fourth and fifth. Note that Bessie cannot run on the fifth minute because she would not end with a rest factor of 0.

解答

题意:Bessie想成为一名运动牛，它每分钟跑的距离为，初始时疲劳度为0，每跑一分钟它就增加1疲劳度，但是它的疲劳度不能超过M。当它选择休息时，它的疲劳度每分钟减少1，但是它必须休息到疲劳度为0时才能继续跑。当N分钟过去后，它的疲劳度必须为0，否则它就没有足够的能量维持生活。求Bessie能跑的最远距离。
思路:定义为第分钟疲劳度为时能够跑的最远距离，若第分钟疲劳度不为，说明上一分钟在跑步，所以；
若第分钟疲劳度为，那么上一分钟疲劳度可能也为，或者歇息了分钟，由此可得以下状态转移方程：
Accepted Code:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10005;
typedef long long ll;
int dp[MAXN][505], a[MAXN];
int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        dp[i][0] = dp[i - 1][0];
        for (int j = 1; j <= m; j++) {
            if (j < i)
                dp[i][0] = max(dp[i][0], dp[i - j][j]);
            dp[i][j] = dp[i - 1][j - 1] + a[i];
        }
    }
    printf("%d\n", dp[n][0]);
    return 0;
}

来源：子夜葵