cf#3A Shortest path of the kinghttp://codeforces.com/contest/3/problem/A

http://codeforces.com/contest/3/problem/A

time limit per test
1 second
memory limit per test
64 megabytes
input
standard input
output
standard output

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.

cf#3A Shortest path of the kinghttp://codeforces.com/contest/3/problem/A_第1张图片

In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input

The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1to 8.

Output

In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: LRUDLULDRU or RD.

LRUD stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

Sample test(s)
input
a8
h1
output
7
RD
RD
RD
RD
RD
RD
RD

A点走到B点,最短路,my code 

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
using namespace std;

int main()
{
    char st[5],ed[5];
    int r1,r2,c1,c2;
    while(cin>>st>>ed)
    {

    r1=st[1]-'0';
    r2=ed[1]-'0';
    c1=st[0]-'a'+1;
    c2=ed[0]-'a'+1;
    if(fabs(r1-r2)==0&&fabs(c2-c1)==0)
    {
        cout<<0<<endl;
        continue;
    }
    if(fabs(r1-r2)>=fabs(c2-c1))
        cout<<fabs(r1-r2)<<endl;
    else
        cout<<fabs(c2-c1)<<endl;
    if(r1==r2)
    {
        if(c1<c2)
        {
            for(int i=0;i<c2-c1;i++)
                cout<<"R"<<endl;
        }
        else
        for(int i=0;i<c1-c2;i++)
        cout<<"L"<<endl;
    }
    else if(c1==c2)
    {
        if(r1<r2)
        {
            for(int i=0;i<r2-r1;i++)
                cout<<"U"<<endl;
        }
        else
            for(int i=0;i<r1-r2;i++)
            cout<<"D"<<endl;
    }
   else if(r1<r2&&c1>c2)///LU
   {
       if(r2-r1>=c1-c2)
       {
           for(int i=0;i<c1-c2;i++)
            cout<<"LU"<<endl;
           for(int i=0;i<r2-r1-c1+c2;i++)
            cout<<"U"<<endl;
       }
       else
       {
            for(int i=0;i<r2-r1;i++)
            cout<<"LU"<<endl;
           for(int i=0;i<c1-c2-r2+r1;i++)
            cout<<"L"<<endl;
       }
   }
   else if(r1<r2&&c1<c2)///RU
   {
       if(r2-r1>=c2-c1)
       {
           for(int i=0;i<c2-c1;i++)
            cout<<"RU"<<endl;
           for(int i=0;i<r2-r1+c1-c2;i++)
            cout<<"U"<<endl;
       }
       else
       {
           for(int i=0;i<r2-r1;i++)
            cout<<"RU"<<endl;
           for(int i=0;i<c2-c1-r2+r1;i++)
            cout<<"R"<<endl;
       }
   }
   else if(r1>r2&&c1>c2)///LD
   {

       if(r1-r2>=c1-c2)
       {
           for(int i=0;i<c1-c2;i++)
            cout<<"LD"<<endl;
           for(int i=0;i<r1-r2-c1+c2;i++)
            cout<<"D"<<endl;
       }
       else
       {
           for(int i=0;i<r1-r2;i++)
            cout<<"LD"<<endl;
            for(int i=0;i<c1-c2-r1+r2;i++)
            cout<<"L"<<endl;
       }

   }
   else///RD r1>r2 c1<c2
   {
       if(r1-r2>=c2-c1)
       {
           for(int i=0;i<c2-c1;i++)
            cout<<"RD"<<endl;
           for(int i=0;i<r1-r2-c2+c1;i++)
            cout<<"D"<<endl;
       }
       else
       {
           for(int i=0;i<r1-r2;i++)
            cout<<"RD"<<endl;
           for(int i=0;i<c2-c1-r1+r2;i++)
           {
               cout<<"R"<<endl;
           }
       }
   }
    }

    return 0;
}





















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