Codeforces 3A-Shortest path of the king（BFS打印路径）

A. 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

a8
h1

output

7
RD
RD
RD
RD
RD
RD
RD

题意：8*8的棋盘 给出起点和终点，求最短路径。
能够走8个方向。

对于打印路径.对一个出队节点x进行搜索的时候将x的每个未訪问子节点的前驱记为x并入队。这样就能通过訪问前驱数组得到逆序的路径.

#include <cstring>
#include <cstdio>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
struct node
{
	int x,y,step;
}path[10][10];
char s[5],e[5];
bool vis[10][10];
int dir[8][2]={{0,1},{0,-1},{-1,0},{1,0},{-1,1},{-1,-1},{1,1},{1,-1}};
void bfs()
{
	queue <node> Q;
	node  t,f;
	t.x=s[0]-'a'+1;t.y=s[1]-'0';t.step=0;
	vis[t.x][t.y]=1;
	Q.push(t);
	while(!Q.empty())
	{
		f=Q.front();Q.pop();
		if(f.x==(e[0]-'a'+1)&&f.y==(e[1]-'0'))
		{
			printf("%d
",f.step);
			stack <int> S;
			while(path[f.x][f.y].x!=-1&&path[f.x][f.y].y!=-1)
			{
				S.push(path[f.x][f.y].step);
				f.x=path[f.x][f.y].x;
				f.y=path[f.x][f.y].y;
			}
			while(!S.empty())
			{
				int tem=S.top();
				S.pop();
				switch (tem)
				{
					case 0: puts("U");break;
					case 1: puts("D");break;
					case 2: puts("L");break;
					case 3: puts("R");break;
					case 4: puts("LU");break;
					case 5: puts("LD");break;
					case 6: puts("RU");break;
					case 7: puts("RD");break;
				}
			}
			return ;
		}
		for(int i=0;i<8;i++)
		{
			t.x=f.x+dir[i][0];
			t.y=f.y+dir[i][1];
			if(t.x>=1&&t.x<=8&&t.y>=1&&t.y<=8&&!vis[t.x][t.y])
			{
				path[t.x][t.y].x=f.x;
				path[t.x][t.y].y=f.y;
				path[t.x][t.y].step=i;
				t.step=f.step+1;
				vis[t.x][t.y]=1;
				Q.push(t);
			}
		}
	}
}
int main()
{
	while(scanf("%s %s",s,e)!=EOF)
	{
		memset(path,-1,sizeof(path));
		memset(vis,0,sizeof(vis));
		bfs();
	}
	return 0;
}