poj2243-Knight Moves-水bfs

#include<iostream>

#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;
#define INF 0x7f
struct node{
   int x, y;
   int cont;
};


bool inq[8][8];
node cb[9][9];
char s1[3],s2[3];
int sx, sy, ex, ey, dx, dy;

int d[8][2]= {{1,2},{1,-2},{2,1},{2,-1},{-1,2},{-1,-2},{-2,1},{-2,-1}};

void init(){
	for(int i = 0; i <= 8; i++)
		for(int j = 0; j <= 8; j++){
			cb[i][j].x = i;
			cb[i][j].y = j;
			cb[i][j].cont = 0;
		}
}

int change(char w){
 int b;
	if(w == 'a') b = 1;
	if(w == 'b') b = 2;
	if(w == 'c') b = 3;
	if(w == 'd') b = 4;
	if(w == 'e') b = 5;
	if(w == 'f') b = 6;
	if(w == 'g') b = 7;
	if(w == 'h') b = 8;
	return b;
}
bool check(int x, int y){
  if(x > 0&&x <= 8&&y > 0&&y <= 8&&!inq[x][y]){
    return true;
  }
  else return false;
}

void bfs(){
	memset(inq, false, sizeof(inq));
	queue<node> q;
    init();
	q.push(cb[sx][sy]);
	node rec;
	inq[sx][sy] = true;
	while(!q.empty()){
		rec = q.front();
		q.pop();
		if(rec.x == ex&&rec.y == ey) break;
		for(int i = 0; i < 8; i++){
            dx = rec.x + d[i][0];
            dy = rec.y + d[i][1];
            if(check(dx,dy)) {
               inq[dx][dy] = true;
               cb[dx][dy].cont = rec.cont+1;
               q.push(cb[dx][dy]);
            }
		}
	}
}

int main(){

	while(scanf("%s%s",&s1,&s2) != EOF){
		sx = change(s1[0]);
		ex = change(s2[0]);

		sy = s1[1] - '0';
		ey = s2[1] - '0';
		bfs();

   printf("To get from %s to %s takes %d knight moves.\n",s1,s2,cb[ex][ey].cont);
	}
	return 0;
}

水的模板BFS，题意就不多解释了，只要知道“马走日”这一规则就可以了，一定会走到终点。

写完后我才发现我逗比地写了一个change()函数，T^T算了，今天逗比了好多次。

末，多谢小建建耐心地帮忙找bug~~~T^T