POJ 3661 Running（区间dp）_poj-3661-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/43908619

本文深入探讨了深度学习和人工智能领域的关键算法和技术，包括卷积神经网络、循环神经网络、支持向量机等，并阐述了这些技术在实际应用中的作用与价值。

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Language:

Running

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5300		Accepted: 1974

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

dp[i][j]表示 i 时刻 j 的疲劳度走的最远距离

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define N 10005

int dp[N][505],n,m,a[N];

int main()
{
	int i,j;
	while(~scanf("%d%d",&n,&m))
	{
		for(i=1;i<=n;i++)
			scanf("%d",&a[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]+a[i];

            dp[i][0]=dp[i-1][0];

			for(int k=1;k<=m&&i>=k;k++)
				dp[i][0]=max(dp[i][0],dp[i-k][k]);
		}
		printf("%d\n",dp[n][0]);
	}
	return 0;
}