阿里巴巴5月5日综合算法题详解 - lazycoding - 博客园


阿里巴巴5月5日综合算法题详解 - lazycoding - 博客园
这道题的大意是:有一个淘宝商户,在某城市有n个仓库,每个仓库的储货量不同,现在要通过货物运输,将每次仓库的储货量变成一致的,n个仓库之间的运输线路围城一个圈,即1->2->3->4->...->n->1->...,货物只能通过连接的仓库运输,设计最小的运送成本(运货量*路程)达到淘宝商户的要求,并写出代码。
解题思路:
假设n个仓库的初始储货量分别为warehouse[1],warehouse[2],...,warehouse[n]
计算平均储货量
average = (warehouse[1]+warehouse[2]+...+warehouse[n])/n
就算出来了最终的结果中,每个仓库应该有的存量
首先,从仓库1向仓库n运送k;
然后,从1到n-1,依次向下运送某一特定值,使得每一个仓库的余量都为average,剩下的问题就是求总代价的最小值了。
设第0步从1仓库向n仓库(注意因为是圆圈,所以路径长度是1)运出k存量,k可以为负,如果为负数,意味着从n向1运输|k|存量,然后从循环,从(1到n-1),从i仓库向i+1仓库运输,运输的量需要保证i仓库在运输完毕后等于average
  1. 第0步(从仓库1向仓库n运送k):花费代价为 |k|,
  2. 第1步(确保仓库1的余量为average):需要花费代价为
    |warehouse[1]-average-k|
    也就是从1向2伙从2向1运输
  3. 第2步(确保仓库2的余量为average):代价为
    |warehouse[2]+warehouse[1]-average-k-average|=|warehouse[1]+warehouse[2]-2*average-k|
...
n-1.第n-1步:代价为
|warehouse[1]+warehouse[2]+...+warehouse[n-1]-(n-1)*average-k|
此时,仓库n剩下的货物量:
(warehouse[n]+k)+warehouse[1]+warehouse[2]+...+warehouse[n-1]-(n-1)*average-k=(warehouse[1]+warehouse[2]+...+warehouse[n])-(n-1)*average=average
刚好也满足,其实这里不用推导,因为平均值是算好的,所以最胡一定是刚好完成的。
总的代价为
|k|+|warehouse[1]-average-k|+|warehouse[1]+a[2]-2*average-k|+...+|warehouse[1]+warehouse[2]+...+warehouse[n-1]-(n-1)*average-k|
不妨令
sum[i] = warehouse[1]+warehouse[2]+...+warehouse[i]-i*average

则,总代价可表示为:
|k|+|sum[1]-k|+|sum[2]-k|+...+|sum[n-1]-k|

这个式子可以看成在水平数轴上寻找一个点k,使得点k到点0,sum[1],sum[2],sum[3],...,sum[n-1]的距离之和最小,显然k应该取这n个数的中位数。至此问题解决。

const int X = 100000;
double sum[X],warehouse[X];
int n;

double Abs(double x)
{
    return max(x,-x);
}

int _tmain(int argc, _TCHAR* argv[])
{
    while(true)
    {
        double total = 0;
        double mid=0;
        cout<<"请输入仓库数目:";
        cin>>n;
        //读入n个仓库的值,并计算总数
        for(int i=1;i<=n;i++)
        {
            cout<<"请输入第"<<i<<"个仓库的存量:";
            cin>>warehouse[i];
            total += warehouse[i];
        }
        //计算每个仓库应该最终存储的值
        double average = total/n;
        //计算sum数组
        for(int i=1;i<n;i++)
            sum[i] = warehouse[i]+sum[i-1]-average;
        //排序后打算去中位数
        //sort采用半开半闭区间,所以排序为0~n-1
        sort(sum,sum+n);
        //这个可以自己举个数字就知道了
        if(n%2!=0)
        {    
            mid = sum[n/2];
        }
        else
        {
            mid=(sum[n/2]+sum[n/2-1])/2;
        }
        cout<<"应该从1开始,运输"<<mid<<"货物,然后依次保证符合条件即可"<<endl;
        double ans = Abs(mid);
        for(int i=1;i<n;i++)
            ans += Abs(sum[i]-mid);
        cout<<"总成本花费是:"<<ans<<endl;
        cout<<"----------------------------------------------------------------------"<<endl;
    }
    return 0;
}

列方程, 将所有的cost用一个变量表示。

av表示平均下来每个仓库的容量。 ki表示仓库i向仓库i+1输送多少(可以为负),列出方程组
wi表示仓库i的存储量
    av=w1+kn-k1
    av=w2+k1-k2
    av=w3+k2-k3
       ...
    av=wn+ kn-1 + kn
将所有k都用kn表示, 有
    k1=kn+ w1 -  av
    k2=kn+ w1+w2 - 2*av
    k3=kn+ w1+w2+w3 - 3*av
        ...
    kn = kn //这里就不必也不能推出别的式子了
令 sum(i) = -(w1+w2+...+wi-i*av) (写成这样方便)
目标函数:
最小一乘, 中位数 sum(0)~sum(n-1)的中位数。{ |kn-sum(1)|+|kn-sum(2)|+...+|kn-sum(n-1)| + |Kn| }
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