| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9978 | Accepted: 4839 |
Description
Each minute, one of the two apple trees drops an apple. Bessie, having much practice, can catch an apple if she is standing under a tree from which one falls. While Bessie can walk between the two trees quickly (in much less than a minute), she can stand under only one tree at any time. Moreover, cows do not get a lot of exercise, so she is not willing to walk back and forth between the trees endlessly (and thus misses some apples).
Apples fall (one each minute) for T (1 <= T <= 1,000) minutes. Bessie is willing to walk back and forth at most W (1 <= W <= 30) times. Given which tree will drop an apple each minute, determine the maximum number of apples which Bessie can catch. Bessie starts at tree 1.
Input
* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.
Output
Sample Input
7 2
2
1
1
2
2
1
1
Sample Output
6
Hint
Seven apples fall - one from tree 2, then two in a row from tree 1, then two in a row from tree 2, then two in a row from tree 1. Bessie is willing to walk from one tree to the other twice.
OUTPUT DETAILS:
Bessie can catch six apples by staying under tree 1 until the first two have dropped, then moving to tree 2 for the next two, then returning back to tree 1 for the final two.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
int i,j,t,n,Max=;
int dp[+][];
int a[+];
freopen("in.txt","r",stdin);
scanf("%d%d",&t,&n);
for(i=;i<=t;i++)
scanf("%d",&a[i]);
for(i=;i<=t;i++)
{
dp[i][]=dp[i-][]+-a[i]; //dp[i][j]={第i分钟走了j步所摘到的苹果}
for(j=;j<=n;j++)
{
if(j%) //如果为奇数，说明走到了2树
dp[i][j]=max(dp[i-][j],dp[i-][j-])+a[i]-; //取前一分钟的最大值
else
dp[i][j]=max(dp[i-][j],dp[i-][j-])+-a[i];
}
}
for(i=;i<=n;i++)
Max=max(dp[t][i],Max);
printf("%d\n",Max); }