POJ3661running

最新推荐文章于 2022-04-27 18:52:45 发布

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在限定疲劳度的情况下，Bessie如何通过合理安排跑步与休息的时间来达到最大的跑步距离。此问题通过动态规划求解最大跑步距离。

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Running

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6818		Accepted: 2532

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

题目大意:N个数代表每隔一秒可以跑的距离，每选择跑一秒意味着疲劳度增加1，休息一秒代表疲劳度减少1，同时要求疲劳度不能超过M，且当跑完后要保证疲劳度为0，问最远能跑多少

解题思路:这是一道DP，就目前我战五渣的能力，基本上能想到DP，但是转化方程什么的基本上想不到怎么去处理，然后看了看题解发现了，b[i][j]中代表第i秒我有j的疲劳度，要想疲劳度为0，就有两种情况，一个是b[i][0]=b[i-1][0]，一个是b[i][0]=b[i-j][j]然后取最大值，同理，对于b[i][j]来说就是b[i-1][j-1]+a[i]

#include<iostream>  
#include<cstdio>
#include<stdio.h>
#include<cstring>  
#include<cstdio>  
#include<climits>  
#include<cmath> 
#include<vector>
#include <bitset>
#include<algorithm>  
#include <queue>
#include<map>
using namespace std;

long long int b[10005][505], a[10005], n, m, i, j;
int main()
{
	cin >> n >> m;
	for (i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	memset(b, 0, sizeof(b));
	for (i = 1; i <= n; i++)
	{
		b[i][0] = b[i - 1][0];
		for (j = 1; j <= m; j++)
		{
			if (i - j >= 0)
			{
				b[i][0] = max(b[i - j][j], b[i][0]);
			}
			b[i][j] = b[i-1][j - 1] + a[i];
		}
	}
	cout << b[n][0] << endl;
}