poj 3661 runnning 区间dp

最新推荐文章于 2019-08-03 11:55:32 发布

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本文介绍了一道关于赛跑的算法题目，目标是通过合理的休息和奔跑策略来最大化跑步距离，同时确保疲劳值最终归零。文章使用了动态规划方法进行解答，并详细展示了代码实现。

Running

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7243		Accepted: 2724

Description

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 D_i (1 ≤ D_i ≤ 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: D_i

Output

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

Sample Input

Sample Output

题意：给你一个n,m，n表示有n分钟，每i分钟对应的是第i分钟能跑的距离，m代表最大疲劳度，每跑一分钟疲劳度+1，当疲劳度==m，必须休息，在任意时刻都可以选择休息，如果选择休息，那么必须休息到疲劳度为0，当疲劳度为0的时候也是可以继续选择休息的，求在n分钟后疲劳度为0所跑的最大距离

dp[i][j]表示第i分钟疲劳值为j时的最大值。

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <string.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 10050;
const int maxm = 550;
int dp[maxn][maxm], d[maxn];
int n, m;
int main() {
    //freopen("in.txt", "r", stdin);
    memset(dp, 0, sizeof(dp));
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        cin >> d[i];
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= min(m, i); ++j) {
            dp[i][j] = dp[i - 1][j - 1] + d[i];
        }
        //疲劳值为0时继续休息
        dp[i][0] = dp[i - 1][0];
        //休息就必须休息到疲劳值为0为止
        for (int j = 1; j <= m; ++j) {
            if (i >= j)
                dp[i][0] = max(dp[i][0], dp[i - j][j]);
        }
    }
    cout << dp[n][0] << endl;
}