poj2385

Apple Catching

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12979 Accepted: 6308

Description

It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveniently numbered 1 and 2) in his field, each full of apples. Bessie cannot reach the apples when they are on the tree, so she must wait for them to fall. However, she must catch them in the air since the apples bruise when they hit the ground (and no one wants to eat bruised apples). Bessie is a quick eater, so an apple she does catch is eaten in just a few seconds. 

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

* Line 1: Two space separated integers: T and W 

* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.

Output

* Line 1: The maximum number of apples Bessie can catch without walking more than W times.

Sample Input

7 22112211

Sample Output

6

题意：一开始在第一棵树下，然后在m分钟内捡苹果，只允许走n次，问最多能捡到多少苹果。

思路：dp[i][j]代表的第i分钟走了j次获得的苹果数，因为一开始是在树1下，根据第一分钟树的不同初始化也有些不一样，然后可以推出递推关系dp[i][j]=max(dp[i-1][j],dp[i][j-1])。

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int dp[1005][35];int main(){    int m,n;    while(scanf("%d%d",&m,&n)!=EOF&&m+n)    {        int i,j;        int a[1005];        for(i=1;i<=m;i++)        {            scanf("%d",&a[i]);        }        memset(dp,0,sizeof(dp));        if(a[1]==1)        {            dp[1][0]=1;            dp[1][1]=0;        }        else        {            dp[1][0]=0;            dp[1][1]=1;        }        int maxn=0;        for(i=2;i<=m;i++)        {            for(j=0;j<=n;j++)            {                if(j==0)dp[i][j]=dp[i-1][j]+a[i]%2;                else                {                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);                    if(j%2+1==a[i])dp[i][j]++;                }                maxn=max(dp[i][j],maxn);            }        }        cout<<maxn<<endl;    }    return 0;}