Poj 3661 running

Transfer

 

Running

Time limit:1000 ms		Memory limit:65536 K

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 factorization ', which starts at 0. When she chooses to run in minuteI, She will run exactly a distanceDi(1 ≤Di≤ 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 ofNMinute 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:NAndM
* Lines 2 ..N+ 1: LineI+ 1 contains the single INTEGER:Di

Output

* Line 1: A single integer representing the largest distance Bessie can run while satisfying the conditions.

Sample Input

5 2534210

Sample output

9

Source

Usaco 2008 January silver feels that if a problem-solving report is written, no question is written .. I am sorry for the number of people who have searched for the question .. The question is that you have m physical strength, and n minutes, for the I minute, you can run dis_ I meters, but the physical strength will be reduced by 1. You can also choose to rest, but if you rest, you can only wait until all your physical strength is restored before you can continue to run the state equation is very clear, DP [I] [J] indicates that the I-th minute consumes J's physical strength. The transfer should be: DP [I + 1] [J + 1] = DP [I] [J] + d [I] // If the DP [I + 1] [0] = DP [I + J] [0] = DP [I] [0], if the answer to the I-minute break is DP [n + 1] [0],

 1 #include<set> 2 #include<queue> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 const int M = 510;10 const int N = 10010;11 #define For(i,n) for(int i=1;i<=n;i++)12 #define Rep(i,l,r) for(int i=l;i<=r;i++)13 14 int n,m,dp[N][M],d[N];15 16 void DP(){17     For(i,n){18         Rep(j,0,m){19             if(j==0) dp[i+1][j]=max(dp[i+1][j],dp[i][j]);20             if(j<m) dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]+d[i]);21             if(i+j<=n+1)   dp[i+j][0]=max(dp[i+j][0],dp[i][j]);22         }23     }24     cout<<dp[n+1][0]<<endl;25 }26 27 int main(){28     scanf("%d%d",&n,&m);29     For(i,n) scanf("%d",&d[i]);30     DP();31     return 0;32 }

 

Poj 3661 running