Knight Moves
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13433 | Accepted: 7518 |
Description
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves.
Source
题意：求点xx到点yy的最少步数
bfs模板
今天学弟问我bfs怎么敲，于是就敲了一遍模板题
stl版
#include<queue> #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=200; int n,m; int vist[N][N]; int dx[8]={-2,-2,-1,-1,1,1,2,2}; int dy[8]={-1,1,-2,2,-2,2,-1,1}; struct node{ int x,y,num; char c; }a,b; queue<node>q; void bfs(){ vist[a.x][a.y]=1; while(!q.empty()){ node temp=q.front();q.pop(); if(temp.x==b.x&&temp.y==b.y){ printf("To get from %c%d to %c%d takes %d knight moves. ",a.c,a.y,b.c,b.y,temp.num); return ; } for(int i=0;i<8;i++){ int xx=temp.x+dx[i]; int yy=temp.y+dy[i]; if(xx<1||xx>8||yy<1||yy>8||vist[xx][yy]) continue; node next; next.x=xx; next.y=yy; next.num=temp.num+1; vist[xx][yy]=1; q.push(next); } } } int main(){ while(~scanf("%c%d %c%d",&a.c,&a.y,&b.c,&b.y)){ getchar(); while(!q.empty()) q.pop(); memset(vist,0,sizeof(vist)); a.x=a.c-'a'+1; b.x=b.c-'a'+1; q.push(a); bfs(); } return 0; }
手工版
#include<iostream> #include<cstdio> #include<cstring> using namespace std; #define N 510 int ax[8]={-2,-2,-1,-1,1,1,2,2}; int ay[8]={-1,1,-2,2,-2,2,-1,1}; int n,m,sx,sy,ex,ey,step[N<<2],dx[N<<2],dy[N<<2]; bool vis[N][N]; char c1[3],c2[3]; void bfs(){ int head=0,tail=1; dx[1]=sx; dy[1]=sy; vis[sx][sy]=1; while(head<tail){ head++; int x0=dx[head],y0=dy[head]; for(int i=0;i<8;i++){ int x1=x0+ax[i],y1=y0+ay[i]; if(!vis[x1][y1]&&x1>0&&x1<=8&&y1>0&&y1<=8) { vis[x1][y1]=1; dx[++tail]=x1;dy[tail]=y1; step[tail]=step[head]+1; //qx[dx[tail]]=dx[head];qy[dy[tail]]=dy[head];//记录前驱->倒序输出 if(x1==ex&&y1==ey){ printf("To get from %s to %s takes %d knight moves. ",c1,c2,step[tail]); return ; } } } } } int main(){ while(scanf("%s%s",c1,c2)==2){ memset(vis,0,sizeof vis); memset(step,0,sizeof step); sx=c1[0]-'a'+1; sy=c1[1]-'0'; ex=c2[0]-'a'+1; ey=c2[1]-'0'; if(sx==ex&&sy==ey){printf("To get from %s to %s takes %d knight moves. ",c1,c2,0);continue;} bfs(); } return 0; }