POJ - 3661 Running
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
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: Di
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
屬於動態規劃,dp【i】【j】代表前i分鐘,疲勞值是j所能跑的最大長度。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 6 using namespace std; 7 8 9 int dp[10005][505],n,m,a[10005]; 10 11 int main() 12 { 13 int i,j; 14 scanf("%d%d",&n,&m); 15 for(i=1;i<=n;i++) 16 scanf("%d",&a[i]); 17 memset(dp,0,sizeof(dp)); 18 for(i=1;i<=n;i++) 19 { 20 for(j=1;j<=m;j++) 21 dp[i][j]=dp[i-1][j-1]+a[i]; 22 dp[i][0]=dp[i-1][0]; 23 for(int k=1;k<=m&&i>=k;k++) 24 dp[i][0]=max(dp[i][0],dp[i-k][k]); 25 } 26 printf("%d\n",dp[n][0]); 27 return 0; 28 }
POJ - 3661 Running