poj 2385 Apple Catching (DP)

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
........................................................................................................................................

这个开始的时候没思路，后来仔细想了想，就有些思路了。
ap[2][t+1]:二维数组保存第i（t>=i>=1)秒时下落的一个苹果，下落就存为1，反之存为0.
a[i][j][k]:还剩 i 秒时且还有 j 次可移动时在 k 树下可以得到的最大苹果树。由此可以得到：

a[i][j][k]=max(  a[i-1][j][k] + ap[k][t-i+1] ,  a[i-1][j-1][1-k] + ap[1-k][t-i+1] ) 
然后就是一些初值：

i 为0时：a[i][j][k]=0
j为0时：a[i][j][k]=for( i --> t ) sum ap[k][i]

然后就可以循环递推求结果了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
    int i,j,k;
    int t,w;
    cin>>t>>w;
    int a[t+1][w+1][2];
    int ap[2][t+1],x;
    for(i=1; i<=t; i++)
    {
        scanf("%d",&x);
        if(x==1)
        {
            ap[0][i]=1;
            ap[1][i]=0;
        }
        else
        {
            ap[0][i]=0;
            ap[1][i]=1;
        }
    }
    int m;
    for(i=0; i<=t; i++)
    {
        for(j=0; j<=w; j++)
        {
            for(k=0; k<2; k++)
            {
                if(i==0)a[i][j][k]=0;
                else if(j==0)
                {
                    int s=0;
                    for(m=t; m>t-i; m--)
                    {
                     s+=ap[k][m];
                    }
                    a[i][j][k]=s;
                }
                else {
                a[i][j][k]=max(a[i-1][j][k]+ap[k][t-i+1],a[i-1][j-1][1-k]+ap[1-k][t-i+1]);
                }
            }
        }
    }
   /* for(i=0; i<=t; i++)
    {
        for(j=0; j<=w; j++)
        {
            for(k=0; k<2; k++)
            {
                printf("%d,%d,%d:%d ",i,j,k,a[i][j][k]);
            }
            printf("\n");
        }
        printf("\n");
    }
    */
    printf("%d\n",a[t][w][0]);
    return 0;
}