POJ 3661 Running

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Running

Time Limit: 1000MS		Memory Limit: 65536K

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

Source

USACO 2008 January Silver

 

感觉如果写解题报告不写题意的话。。有些对不起搜题解的人额。。

题目意思是你有m的体力，以及n分钟的时间，

对于第i分钟，你可以跑dis_i米，但体力会减少1。

也可以选择休息，但是如果休息的话，就只能等体力全部恢复完才可以继续跑

那么状态方程就很明确了，dp[i][j]表示第i分钟耗费了j体力。

转移应为：dp[i+1][j+1]=dp[i][j]+d[i]//如果第i分钟跑了

　　          dp[i+1][0]=dp[i+j][0]=dp[i][0],如果第i分钟休息

答案应为dp[n+1][0]

#include<set>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int M = 510;
const int N = 10010;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)

int n,m,dp[N][M],d[N];

void DP(){
    For(i,n){
        Rep(j,0,m){
            if(j==0) dp[i+1][j]=max(dp[i+1][j],dp[i][j]);
            if(j<m) dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]+d[i]);
            if(i+j<=n+1)   dp[i+j][0]=max(dp[i+j][0],dp[i][j]);
        }
    }
    cout<<dp[n+1][0]<<endl;
}

int main(){
    scanf("%d%d",&n,&m);
    For(i,n) scanf("%d",&d[i]);
    DP();
    return 0;
}