poj-3661 Running（DP）

http://poj.org/problem?id=3661

 

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

5 2
5
3
4
2
10

Sample Output

 9

编程思想：

设dp[i][j]表示第i分钟疲劳值为j的最优状态，每分钟都有两种选择，

该分钟选择跑时的最优状态由上一分钟的最优状态决定，状态转移方程为：dp[i][j]=dp[i-1][j-1]+D[i],

该分钟选择休息时，由于一开始休息就要等疲劳值恢复为0时才可以开始选择跑或继续休息，在开始休息到疲劳值还未减为0的这段时间（用k表示）的每一分钟，只能选择休息，故该状态的最优状态由前面开始决定休息的时间点（i-k）决定，则态转移方程为：dp[i][0]=max(dp[i-k][k]);其中1=<k<=i-k。
题解AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
#include<cmath>
#include<vector>
#include<string.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
int dp[10001][501];
int D[10011];
int main()
{
   //freopen("D:\in.txt","r",stdin);
    int T,cas,i,j,k,n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&D[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]+D[i];
            }
            dp[i][0]=dp[i-1][0];
            for(k=1;k+k<=i;k++)
            {
                dp[i][0]=max(dp[i][0],dp[i-k][k]);
            }
        }
        printf("%d
",dp[n][0]);
    }
    return 0;
}

自己AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <queue>
#include <set>
const int INF=0x3f3f3f3f;
using namespace std;
#define maxn 10010

int a[maxn];
int dp[505][maxn];

int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int MAX=0;
    dp[1][1]=a[1];    
    for(int j=2;j<=n;j++)
    { 
        int x=1,y=j-1;
        while(y>0&&x<=m)//找dp[0][j] 
        {
            if(dp[x][y]>dp[0][j])
                dp[0][j]=dp[x][y];
            x++;
            y--;
            if(dp[0][j]>MAX)
                MAX=dp[0][j];
            else dp[0][j]=MAX;
        }
        
        for(int i=1;i<=m;i++)//更新其余的 
        {
            dp[i][j]=dp[i-1][j-1]+a[j];
        }
            
    }
    printf("%d
",dp[0][n]);
    return 0;
}