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 Z http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vciBhIGNvbW1vbiB2ZXJ0ZXggd2l0aCB0aGUgc3F1YXJlIHRoZSBraW5nIGlzIGN1cnJlbnRseSBpbiAoZ2VuZXJhbGx5IHRoZXJlIGFyZSA4IGRpZmZlcmVudCBzcXVhcmVzIGhlIGNhbiBtb3ZlIHRvKS48L3A+CgoKCklucHV0CjxwPgpUaGUgZmlyc3QgbGluZSBjb250YWlucyB0aGUgY2hlc3Nib2FyZCBjb29yZGluYXRlcyBvZiBzcXVhcmUgPGVtPnM8L2VtPiwgdGhlIHNlY29uZCBsaW5lIKGqIG9mIHNxdWFyZSA8ZW0+dDwvZW0+LjwvcD4KPHA+CkNoZXNzYm9hcmQgY29vcmRpbmF0ZXMgY29uc2lzdCBvZiB0d28gY2hhcmFjdGVycywgdGhlIGZpcnN0IG9uZSBpcyBhIGxvd2VyY2FzZSBMYXRpbiBsZXR0ZXIgKGZyb20gYSB0byBoKSwKIHRoZSBzZWNvbmQgb25lIGlzIGEgZGlnaXQgZnJvbSAxIHRvIDguPC9wPgoKCgpPdXRwdXQKPHA+CkluIHRoZSBmaXJzdCBsaW5lIHByaW50IDxlbT5uPC9lbT4goaogbWluaW11bSBudW1iZXIgb2YgdGhlIGtpbmc="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) inputa8 h1output
7 RD RD RD RD RD RD RD
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#include #include #include #define M 10 using 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 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; }