动态规划刷题

分金子（360公司2017春招真题）

题目链接

https://exercise.acmcoder.com/online/online_judge_ques?ques_id=3863&konwledgeId=42
可以去链接里看一看，不再赘述题目了。

解题思路

假设几个变量，f(i,j)是从[i，j]中可以拿到的最大价值，因为只可以从两端拿，假如先手是从最左端拿的，那剩下的最大价值就是f(i+1,j)，也就是后手能拿到的最大价值；那先手能拿到的最大价值可以用sum(i,j)-f(i+1,j)来表示。
假如先手是从最右端拿的，剩下最大价值就是f(i,j-1)，也就是后手能拿到的最大价值；那先手能拿到的最大价值可以用sum(i,j)-f(i,j-1)来表示。
那究竟应该从哪边拿，当然是看f(i+1,j)和f(i,j-1)的大小，最后可以提炼出

f(i,j) = max{(sum(i,j)-f(i+1,j)),sum(i,j)-f(i,j-1)} = sum(i,j)-min{f(i+1,j),f(i,j-1)}

代码实现

递归版

从上式可以看出，递归是一种解决方案，为了避免重复计算，使用一个变量来记录算出来的f(n,m)的值。

def digui(lst,i,j):
    if(i==j):
        return lst[i]
    elif(res[i][j]!=0):
        return res[i][j]
    else:
        res[i][j]=sum(lst[i:j+1])-min(digui(lst,i+1,j),digui(lst,i,j-1))
        return res[i][j]

n=int(input())
for q in range(n):
    num=int(input())
    res = [[0]*(num+1) for i in range(num+1)]
    lst=list(map(int,input().split()))
    lst=[0]+lst
    a=digui(lst,1,num)
    b=sum(lst)-digui(lst,1,num)
    print("Case #%s: %s %s"%(q+1,a,b))

动态规划版

我也说不上来什么叫做动态规划，总之大家都是这样叫的，当初看别人代码时，也是一脸懵比，然后自己整个手推了一遍，豁然开朗。
我们使用给的案例

4 7 2 9 5 2

上面是一列金子，先画一张表，为了方便表示，我们从1开始，0列0行都当作是空白

‘’	0	1	2	3	4	5	6
0	‘’	‘’	‘’	‘’	‘’	‘’	‘’
1	‘’	‘’	‘’	‘’	‘’	‘’	‘’
2	‘’	‘’	‘’	‘’	‘’	‘’	‘’
3	‘’	‘’	‘’	‘’	‘’	‘’	‘’
4	‘’	‘’	‘’	‘’	‘’	‘’	‘’
5	‘’	‘’	‘’	‘’	‘’	‘’	‘’
6	‘’	‘’	‘’	‘’	‘’	‘’	‘’

接下来可以填表了，假如是从[1,1]取，那就一种取法，就是该位置上的数量，其他只有一个金子的情形也是这样，填表。

‘’	0	1	2	3	4	5	6
0	‘’	‘’	‘’	‘’	‘’	‘’	‘’
1	‘’	4	‘’	‘’	‘’	‘’	‘’
2	‘’	‘’	7	‘’	‘’	‘’	‘’
3	‘’	‘’	‘’	2	‘’	‘’	‘’
4	‘’	‘’	‘’	‘’	9	‘’	‘’
5	‘’	‘’	‘’	‘’	‘’	5	‘’
6	‘’	‘’	‘’	‘’	‘’	‘’	2

那如果是[1,2]该怎么取呢，公式交给我们，应该是先得到[1,2]的和，然后减去较小的那个，那[2,3],[3,4]也是类似。

‘’	0	1	2	3	4	5	6
0	‘’	‘’	‘’	‘’	‘’	‘’	‘’
1	‘’	4	7	‘’	‘’	‘’	‘’
2	‘’	‘’	7	7	‘’	‘’	‘’
3	‘’	‘’	‘’	2	9	‘’	‘’
4	‘’	‘’	‘’	‘’	9	9	‘’
5	‘’	‘’	‘’	‘’	‘’	5	5
6	‘’	‘’	‘’	‘’	‘’	‘’	2

规律很明显了，接着填表即可，最后表的状态如下。

‘’	0	1	2	3	4	5	6
0	‘’	‘’	‘’	‘’	‘’	‘’	‘’
1	‘’	4	7	6	16	11	18
2	‘’	‘’	7	7	11	16	24
3	‘’	‘’	‘’	2	9	7	11
4	‘’	‘’	‘’	‘’	9	9	11
5	‘’	‘’	‘’	‘’	‘’	5	5
6	‘’	‘’	‘’	‘’	‘’	‘’	2

那[1,6]最大的收益就是1行6列的18了，后面只需要交给计算机来帮我们填表即可。

n=int(input())
for q in range(n):
    num=int(input())
    lst = list(map(int,input().split()))
    res = [[0]*(num+1) for i in range(num+1)]
    for i in range(1,num+1):
        res[i][i] = lst[i-1]
    cnt=1
    while cnt<=num:
        i=1
        j=i+cnt
        while j<=num:
            res[i][j] = sum(lst[i-1:j]) - min(res[i][j-1],res[i+1][j])
            i+=1
            j+=1
        cnt+=1
    a = res[1][num]
    b = sum(lst) - a
    print("Case #%s: %s %s"%(q+1,a,b))