Shortest path of the king

Shortest path of the king

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.

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: 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

a8h1

output

7RDRDRDRDRDRDRD

一个棋盘，用字母和数字表示横轴和竖轴，可以向八个方向移动，求最短路并写出路径。

最简单的想法就是模拟，首先最短路大小可以判断，就是横向和纵向的最大值，然后根据大小先斜着走，然后在上下左右走。

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int max(int a, int b){return a > b ? a : b;}int  min(int a, int b){return a < b ? a : b;}int main(){int i, j, m, n, ans, pre1, pre2, temp1, temp2, temp;char a[2], b[2];scanf("%s", a);scanf("%s", b);pre1 = b[0] - a[0];pre2 = b[1] - a[1];temp1 = max(pre1, -pre1); temp2 = max(pre2, -pre2);ans = max(temp1, temp2);if (pre1 > 0 && pre2 > 0){cout << ans << endl;temp = min(temp1, temp2);for (i = 1; i <= temp; i++)cout << "RU" << endl;if (temp1>temp2)for (i = temp + 1; i <= temp1; i++)cout << "R" << endl;elsefor (i = temp + 1; i <= temp2; i++)cout << "U" << endl;}else if (pre1 > 0 && pre2 < 0){cout << ans << endl;temp = min(temp1, temp2);for (i = 1; i <= temp; i++)cout << "RD" << endl;if (temp1>temp2)for (i = temp + 1; i <= temp1; i++)cout << "R" << endl;elsefor (i = temp + 1; i <= temp2; i++)cout << "D" << endl;}else if (pre1 < 0 && pre2 < 0){cout << ans << endl;temp = min(temp1, temp2);for (i = 1; i <= temp; i++)cout << "LD" << endl;if (temp1>temp2)for (i = temp + 1; i <= temp1; i++)cout << "L" << endl;elsefor (i = temp + 1; i <= temp2; i++)cout << "D" << endl;}else if (pre1<0 && pre2>0){cout << ans << endl;temp = min(temp1, temp2);for (i = 1; i <= temp; i++)cout << "LU" << endl;if (temp1>temp2)for (i = temp + 1; i <= temp1; i++)cout << "L" << endl;elsefor (i = temp + 1; i <= temp2; i++)cout << "U" << endl;}if (pre1 == 0){cout << ans << endl;if (pre2>0)for (i = 1; i <= temp2; i++)cout << "U" << endl;elsefor (i = 1; i <= temp2; i++)cout << "D" << endl;}else if (pre2 == 0&&pre1!=0){cout << ans << endl;if (pre1>0)for (i = 1; i <= temp1; i++)cout << "R" << endl;elsefor (i = 1; i <= temp1; i++)cout << "L" << endl;}return 0;}

但是这个方法太麻烦了，写了很多行，并不是题目的最优解。看了一下源代码，发现可以模拟这个人走的坐标，即每次移动改变字符串。 代码简单易懂，值得思考。

#include<cstring>#include<string>#include<iostream>#include<cstdio>using namespace std;int max(int a, int b){return a > b ? a : b;}int main(){string a, b;int i, j, m, n, ans, pre1, pre2;cin >> a >> b;pre1 = max(a[0] - b[0], b[0] - a[0]);pre2 = max(a[1] - b[1], b[1] - a[1]);ans = max(pre1, pre2);cout << ans << endl;while (a != b){if (a[0] < b[0]){cout << "R";a[0]++;}if (a[0]>b[0]){cout << "L";a[0]--;}if (a[1] < b[1]){cout << "U";a[1]++;}if (a[1]>b[1]){cout << "D";a[1]--;}cout << endl;}return 0;}