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  • Educational Codeforces Round 84 (Rated for Div. 2) C. Game with Chips

    C. Game with Chips

    time limit per test1 second
    memory limit per test256 megabytes
    inputstandard input
    outputstandard output
    Petya has a rectangular Board of size n×m. Initially, k chips are placed on the board, i-th chip is located in the cell at the intersection of sxi-th row and syi-th column.
    In one action, Petya can move all the chips to the left, right, down or up by 1 cell.
    If the chip was in the (x,y) cell, then after the operation:
    left, its coordinates will be (x,y−1);
    right, its coordinates will be (x,y+1);
    down, its coordinates will be (x+1,y);
    up, its coordinates will be (x−1,y).
    If the chip is located by the wall of the board, and the action chosen by Petya moves it towards the wall, then the chip remains in its current position.
    Note that several chips can be located in the same cell.
    For each chip, Petya chose the position which it should visit. Note that it’s not necessary for a chip to end up in this position.
    Since Petya does not have a lot of free time, he is ready to do no more than 2nm actions.
    You have to find out what actions Petya should do so that each chip visits the position that Petya selected for it at least once. Or determine that it is not possible to do this in 2nm actions.
    Input
    The first line contains three integers n,m,k (1≤n,m,k≤200) — the number of rows and columns of the board and the number of chips, respectively.
    The next k lines contains two integers each sxi,syi (1≤sxi≤n,1≤syi≤m) — the starting position of the i-th chip.
    The next k lines contains two integers each fxi,fyi (1≤fxi≤n,1≤fyi≤m) — the position that the i-chip should visit at least once.
    Output
    In the first line print the number of operations so that each chip visits the position that Petya selected for it at least once.
    In the second line output the sequence of operations. To indicate operations left, right, down, and up, use the characters L,R,D,U respectively.
    If the required sequence does not exist, print -1 in the single line.
    Examples
    inputCopy
    3 3 2
    1 2
    2 1
    3 3
    3 2
    outputCopy
    3
    DRD
    inputCopy
    5 4 3
    3 4
    3 1
    3 3
    5 3
    1 3
    1 4
    outputCopy
    9
    DDLUUUURR
    题意:
    给你 k 个点,告诉你它们的起点和终点,让你找一条路径,使每个点至少经过终点一次,路径长度不超过2nm

    思路:我们可以选择在最右上的点把它移动到(0,0)的位置,也就是左下角,如果它已经到了左下角,那么其他的点也肯定到达左下角,然后我们从(0,0)的位置开始走S型,遍历全图,所以这样每个点都能到达自己需要到达的点,最坏情况也就是(x+y)+n*m<2*n*m//(x,y)是给出的点中最右上的那个点

    Tip:当然我们每次也可以选取(1,m)这个点,也就是最右上的那个位子,这样可以省略找给出的点中最右上的的那个点,这样的长度是 (n+m)+n*m<2*n*m  。

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int n,m,k;
        cin>>n>>m>>k;
        for(int i=0;i<k;i++)
        {
            int x,y;
            cin>>x>>y;
    
        }
        for(int i=0;i<k;i++)
        {
            int x,y;
            cin>>x>>y;
        }
        string s;
        for(int i=n-1;i>=0;i--)
            s+='D';
        for(int i=m-1;i>=0;i--)
            s+='L';
     for(int i=0;i<m;i++)
     {
         if(i%2==0)
         {
             for(int i=0;i<n-1;i++)
             {
                 s+='U';
             }
             if(i<m-1)
             s+='R';
         }
    
         else{
    
             for(int i=0;i<n-1;i++)
             {
                 s+='D';
             }
              if(i<m-1)
             s+='R';
    
         }
    
    
     }
      cout <<s.size()<<endl;
      cout<<s<<endl;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/sszywq/p/12563052.html
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