A. Shortest path of the king

最新推荐文章于 2019-06-07 18:52:52 发布

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本文介绍了一种解决国王在棋盘上从起点到终点最短路径问题的算法。通过判断国王当前位置与目标位置的关系，选择合适的移动策略，实现最少步数到达目标。算法详细展示了在不同情况下国王的移动方向和步骤。

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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.

$\text{[math]}$

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 test(s)

Input

a8
h1

Output

7
RD
RD
RD
RD
RD
RD
RD

//x1,y1为原点 进行四个方向判断 注意：两点之间直线最短。
#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
char x1,x2;
int y1,y2;
string an[20];
int main()
{
    int ans;
    while(scanf("%c%d",&x1,&y1)!=EOF)
    {
        getchar();
        scanf("%c%d",&x2,&y2);
        getchar();
        x1-='a'+1;
        x2-='a'+1;
        ans=-1;
        //讨论四种情况
        // 右上部分
        if(x2>=x1&&y2>=y1)
        {
            if(y1!=y2&&x1!=x2)
            {
                while(y1!=y2&&x1!=x2)
                {
                    ++ans;
                    x1++;
                    y1++;
                    an[ans]="RU";
                }
            }
            if(y1==y2&&x1!=x2)
            {
                while(x1!=x2)
                {
                    ++ans;
                    x1++;
                    an[ans]="R";
                }
            }

            if(y1!=y2&&x1==x2)
            {
                while(y1!=y2)
                {
                    ++ans;
                    y1++;
                    an[ans]="U";
                }
            }
        }
        // 左上部分
        else if(x2<=x1&&y2>=y1)
        {
            if(y1!=y2&&x1!=x2)
            {
                while(y1!=y2&&x1!=x2)
                {
                    ++ans;
                    x1--;
                    y1++;
                    an[ans]="LU";
                }

            }
            if(y1==y2&&x1!=x2)
            {
                while(x1!=x2)
                {
                    ++ans;
                    x1--;
                    an[ans]="L";
                }
            }

            if(y1!=y2&&x1==x2)
            {
                while(y1!=y2)
                {
                    ++ans;
                    y1++;
                    an[ans]="U";
                }
            }
        }
        //左下部分
        else if(x2<=x1&&y2<=y1)
        {
            if(y1!=y2&&x1!=x2)
            {
                while(y1!=y2&&x1!=x2)
                {
                    ++ans;
                    x1--;
                    y1--;
                    an[ans]="LD";
                }
            }
            if(y1==y2&&x1!=x2)
            {
                while(x1!=x2)
                {
                    ++ans;
                    x1--;
                    an[ans]="L";
                }
            }
            if(y1!=y2&&x1==x2)
            {
                while(y1!=y2)
                {
                    ++ans;
                    y1--;
                    an[ans]="D";
                }
            }
        }
        //右下部分
        else if(x2>=x1&&y2<=y1)
        {
            if(y1!=y2&&x1!=x2)
            {
                while(y1!=y2&&x1!=x2)
                {
                    ++ans;
                    x1++;
                    y1--;
                    an[ans]="RD";
                }
            }
            if(y1==y2&&x1!=x2)
            {
                while(x1!=x2)
                {
                    ++ans;
                    x1++;
                    an[ans]="R";
                }
            }
            if(y1!=y2&&x1==x2)
            {
                while(y1!=y2)
                {
                    ++ans;
                    y1--;
                    an[ans]="D";
                }
            }
        }
        printf("%d\n",ans+1);
        for(int i=0; i<=ans; i++)
        {
            printf("%s\n",an[i].c_str());
        }
    }
    return 0;
}