poj2385（动态规划）

Apple Catching

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12945 Accepted: 6294

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：有两棵树，每隔一分钟两棵树中有一颗会掉下苹果，一共T分钟，现在一个人站在树1下，他最多只想在两个树间走W次，走的很快。求他最多能从空中接到到几个苹果。

#include<iostream>#include<stdio.h>  #include<string.h>  #include<math.h>#include<algorithm>using namespace std;  #define maxt 1005#define maxw 35int t,w;int dp[maxw][maxt];int tree[maxt];     //第1分钟掉落苹果的树 int solve(){memset(dp,0,sizeof(dp));dp[(tree[1]-1)%2][1]=1;      //表示第一分钟要接到一颗需要走几次 for(int i=2;i<=t;i++)      //先把走的次数为0的算出来 dp[0][i]=dp[0][i-1]+tree[i]%2; for(int i=1;i<=w;i++){for(int j=2;j<=t;j++){//传递公式，j分钟，走i步，能不能接到当前这个苹果是确定的，所以算出j-1分钟能拿的最多的苹果数就好了 //根据当前是不是再走一次的判断，比较j-1分钟时候走i步和i-1步 dp[i][j]=dp[i][j-1]>dp[i-1][j-1]?dp[i][j-1]:dp[i-1][j-1];   if(tree[j]-1==i%2)dp[i][j]++;}}//由于在这么多时间内走多少次所能接到的苹果最多不确定所以要逐一比较取最大 int res=dp[0][t];for(int i=0;i<=w;i++)if(res<dp[i][t])res=dp[i][t];return res;}int main(){while(cin>>t>>w){for(int i=1;i<=t;i++)cin>>tree[i];cout<<solve()<<endl;}}