[USACO 2008 Jan S]Running

NC24949. [USACO 2008 Jan S]Running

描述

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.

输入描述

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

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:

输出:

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.

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 41ms, 内存消耗: 20068K, 提交时间: 2019-07-28 16:43:33

#include <bits/stdc++.h>
using namespace std;
int dp[10005][505];
int main() {
	int n, m, num;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &num);
		dp[i][0] = dp[i-1][0];
		for (int j = 1; j <= m; j++) {
			if (i > j) dp[i][0] = max(dp[i][0], dp[i-j][j]);
			dp[i][j] = dp[i-1][j-1] + num;
		}
	}
	printf("%d\n", dp[n][0]);
	return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 15ms, 内存消耗: 624K, 提交时间: 2020-03-29 16:33:13

#include<bits/stdc++.h>
using namespace std;
int f[10001],n,m,s,sum[10001],i,k;
int main()
{
	cin>>n>>m;
	for(i=1;i<=n;i++)
	{
		cin>>s;
		sum[i]=sum[i-1]+s;
		f[i]=f[i-1];
		for(k=1;k<=m&&i-k*2>=0;k++)
		f[i]=max(f[i],f[i-k*2]+sum[i-k]-sum[i-k*2]);
	}
	cout<<f[n]<<endl;
}

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