439 - Knight Moves

题目：439 - Knight Moves

题目大意：就是给起点，终点，看最少走几步能从起点到终点。走的方式是采用国际象棋，走日字型。

解题思路：BFS，广度优先遍历。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

struct bit {

	int x, y;
};

const int N = 8;
int dir[8][2] = {{-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}, {1, -2}, {2, -1}, {2, 1}, {1, 2}};
char s1[N], s2[N];
queue<bit> q;
int path[N + 2][N + 2], visit[N + 2][N + 2];

void bfs(bit k) {
	
	visit[k.x][k.y] = 1;
	if(k.x == s2[1] - '0' && k.y == s2[0] - 'a' + 1)
		return;
	q.push(k);
	while(!q.empty()) {
		
		k = q.front();
		q.pop();
		for(int i = 0; i < N; i++) {

			int x = k.x  + dir[i][0];
			int y = k.y  + dir[i][1];
			if(x > 0 && x <= N && y > 0 && y <= N && visit[x][y] == 0 ) {
				
			
				visit[x][y] = 1;	
				path[x][y] = path[k.x][k.y] + 1;	
				if(x == s2[1] - '0' && y == s2[0] - 'a' + 1)
					return;
				bit m;
				m.x = x;
				m.y = y;
				q.push(m); 
			}
		}

	}
	 

}

int main() {
	
	while(scanf("%s%s", s1, s2) != EOF) {
		
		memset(path, 0, sizeof(path));
		memset(visit, 0, sizeof(visit));
	
		bit k;
		k.x = s1[1] - '0';
		k.y = s1[0] - 'a' + 1;
		bfs(k);
		if(visit[s2[1]- '0'][s2[0] - 'a' + 1])
		printf("To get from %s to %s takes %d knight moves.\n", s1, s2, path[s2[1] - '0'][s2[0] - 'a' + 1]);
		
		while(!q.empty()) {

			q.pop();
		}
	}
	return 0;
}