Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8898 | Accepted: 5112 |
Description
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
Output
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
bfs练习题
#include <iostream> #include <stdio.h> #include <queue> #include <vector> #include <string.h> #include <algorithm> using namespace std; struct node { int x,y; }; int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1}; int dis[10][10]; int ex,ey; bool OK(node &b) { if(b.x<1||b.x>8||b.y<1||b.y>8||dis[b.x][b.y]) return false; return true; } void BFS(int x,int y) { node a,b; queue<node> Q; a.x=x;a.y=y; Q.push(a); int i; while(!Q.empty()) { a=Q.front();Q.pop(); for(i=0;i<8;i++) { b.x=a.x+dir[i][0]; b.y=a.y+dir[i][1]; if(b.x==x&&b.y==y) continue; if(OK(b)) { dis[b.x][b.y]=dis[a.x][a.y]+1; Q.push(b); } if(b.x==ex&&b.y==ey) return; } } } int main() { char op1[5],op2[5]; while(scanf("%s %s",op1,op2)!=EOF) { memset(dis,0,sizeof(dis)); int x=op1[0]-'a'+1,y=op1[1]-'0'; ex=op2[0]-'a'+1;ey=op2[1]-'0'; // printf("%d %d==\n",ex,ey); if(x!=ex||y!=ey) BFS(x,y); printf("To get from %s to %s takes %d knight moves.\n",op1,op2,dis[ex][ey]); } }