Codeforces 3A Shortest path of the king(BFS???NO!!)

Shortest path of the king

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  CodeForces. Original ID:  3A
64-bit integer IO format:  %I64d      Java class name:  (Any)

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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.

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 1 to 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: L, R, U, D, LU, LD, RU or RD.

L, R, U, D 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 Input

Input

a8
h1

Output

7
RD
RD
RD
RD
RD
RD
RD

Source

Codeforces Beta Round #3

题意:国际象棋的王从一个点到另外一个点的最短路径的每一步的方向,原来刚开始想到的是BFS,但是一想有点复杂,一步小心又看到了q神的代码,ORZ,竟然可以这么写,遂参考了他的代码.

AC代码:

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    char s1[5],s2[5];
    while(cin>>s1>>s2)
    {
        int x=s2[0]-s1[0];//LR
        int y=s2[1]-s1[1];//UD
        cout<<max(abs(x),abs(y))<<endl;
        while(x||y)
        {
            if(x>0)
                x--,cout<<"R";
            if(x<0)
                x++,cout<<"L";
            if(y>0)
                y--,cout<<"U";
            if(y<0)
                y++,cout<<"D";
            cout<<endl;
        }
    }
    return 0;
}