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    emu in blogjava

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    Problem Statement

    ???? There are three stacks of crates - two of them outside of the warehouse, and one inside the warehouse. We have a crane that can move one crate at a time, and we would like to move all of the crates to the stack inside the warehouse. A heavier crate can never be stacked on top of a lighter crate, and all three initial stacks obey that rule.

    Create a class CraneWork that contains a method moves that is given int[]s stack1, stack2, and warehouse containing the initial three stacks, and returns the minimum number of moves required to move all the crates into the warehouse stack. The elements of stack1, stack2, and warehouse represent the weights of the crates and are given from top to bottom (and thus in non-decreasing order of weight).

    Definition

    ????
    Class: CraneWork
    Method: moves
    Parameters: int[], int[], int[]
    Returns: int
    Method signature: int moves(int[] stack1, int[] stack2, int[] warehouse)
    (be sure your method is public)
    ????

    Constraints

    - stack1, stack2, and warehouse will each contain between 0 and 20 elements, inclusive.
    - The total number of elements in the three stacks will be between 1 and 20, inclusive.
    - Each element in the three stacks will be between 1 and 200, inclusive.
    - stack1, stack2, and warehouse will each be in non-decreasing order.

    Examples

    0)
    ????
    {3,50}
    {}
    {1,2,50,50,50}
    Returns: 12
    Move 3 to stack2, 1 to stack1, 2 to stack2, 1 to stack2, 50 to warehouse, 1 to warehouse, 2 to stack1, 1 to stack1, 3 to warehouse, 1 to stack2, 2 to warehouse, 1 to warehouse.
    1)
    ????
    {50}
    {50}
    {10,20,30}
    Returns: 17
    Start by moving 50 from stack2 to stack1. It then takes 7 moves to transfer the 10,20,30 to stack 2, 2 moves to transfer the 2 50's to the warehouse, and 7 more to transfer the 10,20,30 to the warehouse.
    2)
    ????
    {}
    {}
    {2,5,6,7}
    Returns: 0
    All the crates are already in the warehouse.
    3)
    ????
    {1,2,3}
    {}
    {}
    Returns: 7
    Move 1 from stack1 to warehouse, 2 from stack1 to stack2, 1 from warehouse to stack2, 3 from stack1 to warehouse, 1 from stack2 to stack1, 2 from stack2 to warehouse, 1 from stack1 to warehouse.

    This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

    posted on 2006-09-05 10:21 emu 閱讀(6645) 評論(7)  編輯  收藏 所屬分類: google編程大賽模擬題及入圍賽真題

    評論

    # re: 1000分模擬題CraneWork 2007-06-09 11:02 匿名
    #include <iostream>
    using namespace std;


    class CStack{
    public:
    CStack(){
    memset(data,0,100*sizeof(int));
    index=0;
    }
    virtual~CStack(){}
    int Pop() { sum++; return data[index--];}
    void Push(int D) { data[++index]=D; }
    int Find(int D,int I)
    {
    for(int i=I;i<=index;i++)
    if(data[i]<D)
    return i;
    return index+1;
    }
    int Show(int I)
    {
    if(index==0) return 0;
    if(I>index||I<1)
    return 0;
    return data[I];
    }
    bool Istop(int T)
    {
    if(T==index) return true;
    else return false;
    }
    void Show()
    {
    if(index==0)
    {
    cout<<"空"<<endl;
    return;
    }

    for(int i=index;i>0;i--)
    cout<<data[i]<<" ";
    cout<<endl;
    }
    public:
    static int sum;
    private:
    int data[100];
    int index;
    };
    int CStack::sum=0;

    CStack stack[3];


    void show()
    {
    cout<<"////////////////"<<endl;
    for(int i=0;i<3;i++)
    stack[i].Show();
    cout<<"////////////////"<<endl;
    }
    void CraneWork(int a,int k1,int b,int k2,int c,int k3)
    {
    int x=stack[a].Show(k1);
    int y=stack[b].Show(k2);
    int z=stack[c].Show(k3);
    if(x==0&&y==0) return;

    if(z>=x&&z>=y)
    {
    if(x>=y)
    {
    int k=stack[c].Find(x,k3);
    CraneWork(a,k1+1,c,k,b,k2);
    stack[c].Push(stack[a].Pop());
    show();
    CraneWork(b,k2,a,k1,c,k+1);
    }
    else
    {
    int k=stack[c].Find(y,k3);
    CraneWork(b,k2+1,c,k,a,k1);
    stack[c].Push(stack[b].Pop());
    CraneWork(a,k1,b,k2,c,k+1);
    }
    }
    if(x>z&&x>=y)
    {
    if(x==y)
    {
    if(stack[a].Istop(k1)&&stack[b].Istop(k2))
    {
    stack[a].Push(stack[b].Pop());
    CraneWork(a,k1+2,c,k3,b,k2);
    stack[c].Push(stack[a].Pop());
    stack[c].Push(stack[a].Pop());
    CraneWork(a,k1,b,k2,c,k3+2);
    return;
    }
    }
    CraneWork(a,k1+1,c,k3,b,k2);
    stack[c].Push(stack[a].Pop());
    show();
    CraneWork(a,k1,b,k2,c,k3+1);

    }
    if(y>z&&y>x)
    {
    CraneWork(b,k2+1,c,k3,a,k1);
    stack[c].Push(stack[b].Pop());
    show();
    CraneWork(a,k1,b,k2,c,k3+1);
    }
    }



    int _tmain(int argc, _TCHAR* argv[])
    {
    /*stack[0].Push(50);
    stack[0].Push(3);
    stack[2].Push(50);
    stack[2].Push(50);
    stack[2].Push(50);
    stack[2].Push(2);
    stack[2].Push(1);*/
    /*stack[0].Push(3);
    stack[0].Push(2);
    stack[0].Push(1);*/
    stack[0].Push(50);
    stack[1].Push(50);
    stack[2].Push(30);
    stack[2].Push(20);
    stack[2].Push(10);


    show();
    CraneWork(0,1,1,1,2,1);
    show();
    cout<<"次數為:"<<CStack::sum<<endl;
    return 0;
    }
      回復  更多評論
      

    # re: 1000分模擬題CraneWork [未登錄] 2007-08-06 18:52 zero
    very easy!  回復  更多評論
      

    # re: 1000分模擬題CraneWork [未登錄] 2007-11-11 22:23 y
    g sdfg  回復  更多評論
      

    # re: 1000分模擬題CraneWork 2008-09-19 09:26
    比賽題目都是英文么  回復  更多評論
      

    # re: 1000分模擬題CraneWork 2008-09-22 11:21 hejian
    這是一道漢諾塔的變種題,不知使用進位迭代的方式還有沒有效,有時間可以試一下。  回復  更多評論
      

    # re: 1000分模擬題CraneWork 2008-11-27 17:59 誠人
    一直對google技術佩服!我也要研究一番。  回復  更多評論
      

    # re: 1000分模擬題CraneWork [未登錄] 2010-10-03 23:54 無名
    鑒定完畢,此程序有問題。  回復  更多評論
      

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