cf A. Shortest path of the king 简单存路径bfs

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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 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

a8h1

output

7RDRDRDRDRDRDRD

在一个8*8的迷宫里面，给你起点和终点坐标，让你输出从起点到终点的最短路径。

//30 ms 0 KB#include<stdio.h>#include<string.h>#include<iostream>#include<queue>#define M 10using namespace std;int dir[8][2]= {{0,1},{0,-1},{1,0},{1,-1},{1,1},{-1,0},{-1,1},{-1,-1}};int g[M][M];int s_x,s_y,e_x,e_y;bool vis[M][M];struct sa{    int x,y,step;} path[30],p[9][9];void bfs(){    sa pre;    queue<sa>q;    pre.x=s_x;    pre.y=s_y;    pre.step=0;    q.push(pre);    vis[s_x][s_y]=1;    while(!q.empty())    {        sa now=q.front(),next;        q.pop();        if(now.x==e_x&&now.y==e_y)        {            int ans=now.step;            printf("%d\n",ans);            for(int d=ans; d>=0; d--)            {                path[d]=now;                now=p[now.x][now.y];            }            if(path[1].x==s_x-1)//先判断第一个点和起点之间的距离            {                if(path[1].y==s_y-1)printf("LD\n");                else if(path[1].y==s_y)printf("D\n");                else if(path[1].y==s_y+1)printf("RD\n");            }            else if(path[1].x==s_x)            {                if(path[1].y==s_y-1)printf("L\n");                else if(path[1].y==s_y+1)printf("R\n");            }            else if(path[1].x==s_x+1)            {                if(path[1].y==s_y-1)printf("LU\n");                else if(path[1].y==s_y)printf("U\n");                else if(path[1].y==s_y+1)printf("RU\n");            }            for(int i=2; i<=ans; i++)//判断剩下都点和前面的点之间的关系            {                if(path[i].x==path[i-1].x-1)                {                    if(path[i].y==path[i-1].y-1)printf("LD\n");                    else if(path[i].y==path[i-1].y)printf("D\n");                    else if(path[i].y==path[i-1].y+1)printf("RD\n");                }                else if(path[i].x==path[i-1].x)                {                    if(path[i].y==path[i-1].y-1)printf("L\n");                    else if(path[i].y==path[i-1].y+1)printf("R\n");                }                else if(path[i].x==path[i-1].x+1)                {                    if(path[i].y==path[i-1].y-1)printf("LU\n");                    else if(path[i].y==path[i-1].y)printf("U\n");                    else if(path[i].y==path[i-1].y+1)printf("RU\n");                }            }            break;        }        for(int i=0; i<8; i++)//判断8个方向        {            int xx=now.x+dir[i][0];            int yy=now.y+dir[i][1];            if(xx>=1&&xx<=8&&yy>=1&&yy<=8&&!vis[xx][yy])//如果没有越界            {                vis[xx][yy]=1;                next.x=xx;                next.y=yy;                next.step=now.step+1;                p[next.x][next.y]=now;//存路径                q.push(next);            }        }    }}int main(){    char s1,e1;    while(cin>>s1>>s_x>>e1>>e_x)    {        memset(vis,0,sizeof(vis));        s_y=s1-'a'+1;        e_y=e1-'a'+1;        bfs();    }    return 0;}