poj2385 简单DP

J - 简单dp

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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 2
2
1
1
2
2
1
1

Sample Output

6

Hint

INPUT DETAILS:

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.

题目大意：有两课苹果树，编号为1和2，每一分钟都会有苹果从其中之一落下，刚开始的时候人在苹果树1的下面，这个人最多可以移动w

次,问最终这个人最多可以拿到多少苹果。

思路分析：本问题中，总共有三个变量，一个是当前是第几分钟（i)，另一个是已经移动了多少次(j)，除此之外，还有这个人当前处在哪一棵苹果树

下，但是我们发现，这个人当前在哪一棵苹果树是可以通过移动的次数推算出来的，也就是说现在的变量只有i,j,我们最后要求的就是dp[T][j]的最大值

同时决策就是在当前分钟要不要移动，决策就会引起状态的转移，dp[i][j]=max(dp[i-1][j]，dp[i-1][j-1]),同时对每一步要进行判断，如果当前所在的

苹果树在i分钟刚好掉落苹果，那么dp[i][j]++;

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
const int maxn1=1000+10;
const int maxn2=30+5;
int dp[maxn1][maxn2];//dp[i][j]代表的是前imin，移动j次可以捡到的最大苹果数
int num[maxn1];
const int inf=0xffffff;
int main()
{
     int t,w;
     while(scanf("%d%d",&t,&w)!=EOF)
     {
         memset(dp,0,sizeof(dp));
        for(int i=1;i<=t;i++)
            scanf("%d",&num[i]);
        if(num[1]==1) dp[1][0]=1,dp[1][1]=0;
        else dp[1][0]=0,dp[1][1]=1;
        for(int i=2;i<=t;i++)
        {
            for(int j=0;j<=w;j++)
            {
                if(j==0) dp[i][j]=dp[i-1][j]+(num[i]==1);
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);
                    if(j%2+1==num[i]) dp[i][j]++;
                }
            }
        }
        int ma=-inf;
            for(int j=0;j<=w;j++)
            if(dp[t][j]>ma) ma=dp[t][j];
        cout<<ma<<endl;
     }
    return 0;
}