描述
http://poj.org/problem?id=2385
两棵苹果树,给定一个时间t,1~t每分钟有一棵树掉苹果,牛起始在#1树,最多换w次位置,问最多接到多少苹果.
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10522 | Accepted: 5106 |
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.
Source
分析
用f[i][j][k]表示第i分钟,已经移动了j次,在#k树下的最优解.
注意:
1.有些时候动规写成+的形式比-的形式方便
#include<cstdio>
#include<algorithm>
using std :: max; const int maxt=,maxw=;
int t,w;
int tree[maxt],f[maxt][maxw][]; inline int move(int x) { return x== ? : ; } void solve()
{
int ans=;
for(int i=;i<t;i++)
{
for(int j=;j<=w;j++)
{
for(int k=;k<=;k++)
{
if(k==tree[i+])
{
f[i+][j][k]=max(f[i+][j][k],f[i][j][k]+);
f[i+][j+][move(k)]=max(f[i+][j+][move(k)],f[i][j][k]);
}
else
{
f[i+][j][k]=max(f[i+][j][k],f[i][j][k]);
f[i+][j+][move(k)]=max(f[i+][j+][move(k)],f[i][j][k]+);
}
}
}
}
for(int i=;i<=t;i++)
{
for(int j=;j<=w;j++)
{
for(int k=;k<=;k++)
{
ans=max(ans,f[i][j][k]);
}
}
}
printf("%d\n",ans);
} void init()
{
scanf("%d%d",&t,&w);
for(int i=;i<=t;i++)
{
scanf("%d",tree+i);
}
if(tree[]==) f[][][]=;
else f[][][]=;
} int main()
{
#ifndef ONLINE_JUDGE
freopen("apple.in","r",stdin);
freopen("apple.out","w",stdout);
#endif
init();
solve();
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
#endif
return ;
}