poj 3661 Running(dp,设计状态,)

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1、http://poj.org/problem?id=3661

2、题目大意：

Bessie要在n分钟内跑完路程，刚开始跑步他的疲劳度是0，每跑步一分钟疲劳度就增加1，疲劳度不能超过上限m,达到m就必须休息，休息一分钟疲劳度就减1，一旦休息必须等到疲劳度到达0之后才可以跑步，当然疲劳度是0的时候，也可以继续选择休息，求在满足要求的情况下，Bessie可以跑得最远距离是多少？

首先设dp[i][j]表示第i分钟疲劳度是j的时候获得的最远距离

dp[i][j]=dp[i-1][j-1]+d[i],这是选择在第i分钟跑步的状态

dp[i][0]=dp[i-1][0]，第i分钟仍然在休息，来自于第i-1分钟休息，

那么一直休息，什么时候最大呢

dp[i][0]=max(dp[i][0],dp[i-k][k]) ,(k<=m;i-k>0)

3、Ac代码

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

4、题目：

Running

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4911		Accepted: 1821

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