CSU-ACM2017暑假训练9-区间DP F - Running

F - Running

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 Di (1 ≤ Di ≤ 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: Di

Output

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

Sample Input

5 2 5 3 4 2 10

Sample Output

9

对于这题，有两类，共三种状态，分别为：第一类第一种，“疲劳值未满，继续奔跑”；第二类第一种，“从某时刻开始的休息结束了”；第二类第二种，“尽管之前已经休息了，但还是继续休息”。

I.1:   II.1:   II.2:   dp[i][j]=dp[i−1][j−1]+sample[i]dp[i][0]=dp[i−k][k]dp[i][0]=dp[i−1][0] 其中， dp[i][j] 表示在第 i 分钟，疲劳值为 j 的情况下马所奔跑的最长路程； sample[i] 表示在第 i 分钟内马所能奔跑的路程。 只需要按照上述方程递推到 i=n 即可得到答案。

#include <iostream> #include <cstdio> #include <algorithm> #include <vector> #include <cstring> #include <queue> using namespace std; int sample[10006], dp[10005][503]; int main(){ #ifdef TEST freopen("test.txt", "r", stdin); #endif // TEST int n, m; while(cin >> n >> m){ memset(dp, 0, sizeof(dp)); for(int i = 1; i <= n; i++) scanf("%d", &sample[i]); for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++){ dp[i][j] = dp[i-1][j-1] + sample[i]; } dp[i][0] = dp[i-1][0]; for(int k = 1; k < i && k <= m; k++)//注意k<=m的条件。如不加以限制会超时。 dp[i][0] = max(dp[i][0], dp[i-k][k]); } cout << dp[n][0] << endl; } return 0; }