Running（区间DP）

最新推荐文章于 2020-04-24 12:23:04 发布

林下的码路

最新推荐文章于 2020-04-24 12:23:04 发布

阅读量1k

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划 poj 分治 ACM 文章标签： ACM 算法 dp

本文链接：https://blog.youkuaiyun.com/Enjoying_Science/article/details/49100793

ACM 同时被 3 个专栏收录

521 篇文章

订阅专栏

动态规划

102 篇文章

订阅专栏

poj

73 篇文章

订阅专栏

本文探讨了如何通过动态规划解决Bessie在有限时间内最大化跑步距离的问题，同时确保其疲劳值始终处于允许范围内。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Link：http://poj.org/problem?id=3661

Running

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5670		Accepted: 2133

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

Source

USACO 2008 January Silver

题意：自己应该好好理解清楚，一开始自己理解错了，所以不想描述出来，下次看时再锻炼下自己的读题能力！！！

编程思想：设dp[i][j]表示第i分钟疲劳值为j的最优状态，每分钟都有两种选择，该分钟选择跑时的最优状态由上一分钟的最优状态决定，状态转移方程为：dp[i][j]=dp[i-1][j-1]+D[i],该分钟选择休息时，由于一开始休息就要等疲劳值恢复为0时才可以开始选择跑或继续休息，在开始休息到疲劳值还未减为0的这段时间（用k表示）的每一分钟，只能选择休息，故该状态的最优状态由前面开始决定休息的时间点（i-k）决定，则态转移方程为：dp[i][0]=max(dp[i-k][k]);其中1=<k<=i-k。

AC code：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
#include<cmath>
#include<vector>
#include<string.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
int dp[10001][501];
int D[10011];
int main()
{
   //freopen("D:\\in.txt","r",stdin);
    int T,cas,i,j,k,n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&D[i]);
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
               dp[i][j]=dp[i-1][j-1]+D[i];
            }
            dp[i][0]=dp[i-1][0];
            for(k=1;k+k<=i;k++)
            {
                dp[i][0]=max(dp[i][0],dp[i-k][k]);
            }
        }
        printf("%d\n",dp[n][0]);
    }
    return 0;
}