poj 3661 Running（dp）

题目链接

Running

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5279 Accepted: 1966

Description

The cows are trying to become better athletes, so Bessie is running on a track for exactlyN (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 minutei, 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 exceedM (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

5 2534210

Sample Output

9

题意：跑步，有n分钟可以跑，第i分钟跑的距离为Di。每一分钟可以选择跑与不跑。跑的话疲劳程度+1，疲劳程度初始时为0，不能超过m。不跑的话每分钟疲劳-1，但是要直到疲劳程度减少为0，才可以再跑。n分钟跑完后，疲劳程度必须为0。问最多能跑多少距离？

题解：动态规划问题。用dp[i][j][0,1] 表示第i分钟，疲劳程度为j，0,1表示这分钟跑与没跑，该状态下跑的最远距离。

详情见代码：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#define nn 11000#define inff 0x3fffffffusing namespace std;int n,m;int d[nn];int dp[nn][510][2];int main(){    int i,j;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&d[i]);        }        for(i=0;i<=n;i++)        {            for(j=0;j<=m;j++)                dp[i][j][0]=dp[i][j][1]=-inff;        }        dp[0][0][1]=0;        for(i=0;i<n;i++)        {            for(j=0;j<=m;j++)            {                if(dp[i][j][1]>=0)                {                    if(j+1==m)                        dp[i+1][m][0]=max(dp[i+1][m][0],dp[i][j][1]+d[i+1]);                    else                        dp[i+1][j+1][1]=max(dp[i+1][j+1][1],dp[i][j][1]+d[i+1]);                    if(j>1)                        dp[i+1][j-1][0]=max(dp[i+1][j-1][0],dp[i][j][1]);                    else                        dp[i+1][0][1]=max(dp[i+1][0][1],dp[i][j][1]);                }                if(dp[i][j][0]>=0)                {                    if(j>1)                        dp[i+1][j-1][0]=max(dp[i+1][j-1][0],dp[i][j][0]);                    else                        dp[i+1][0][1]=max(dp[i+1][0][1],dp[i][j][0]);                }            }        }        printf("%d\n",dp[n][0][1]);    }    return 0;}