POJ 2243 ZOJ 1091 UVaOJ 439 Knight moves(BFS廣度優先搜尋)

//典型的BFS（廣度優先搜尋）<br />//馬走日所需最少步數問題<br />#include<iostream><br />using namespace std;<br />int map[8][8];//記錄步數數組<br />int vis[8][8];//訪問過的記錄數組<br />char a[3],b[3];<br />int x1,y1,x2,y2;<br />void set(int x1,int y1)<br />{<br />//剪去不合理範圍<br />if(x1+1 <= 7 && y1+2 <= 7 && !vis[x1+1][y1+2])<br />map[x1+1][y1+2] = map[x1][y1]+1;<br />if(x1+1 <= 7 && y1-2 >= 0 && !vis[x1+1][y1-2])<br />map[x1+1][y1-2] = map[x1][y1]+1;<br />if(x1-1 >= 0 && y1+2 <= 7 && !vis[x1-1][y1+2])<br />map[x1-1][y1+2] = map[x1][y1]+1;<br />if(x1-1 >= 0 && y1-2 >= 0 && !vis[x1-1][y1-2])<br />map[x1-1][y1-2] = map[x1][y1]+1;<br />if(x1+2 <= 7 && y1+1 <= 7 && !vis[x1+2][y1+1])<br />map[x1+2][y1+1] = map[x1][y1]+1;<br />if(x1-2 >= 0 && y1+1 <= 7 && !vis[x1-2][y1+1])<br />map[x1-2][y1+1] = map[x1][y1]+1;<br />if(x1+2 <= 7 && y1-1 >= 0 && !vis[x1+2][y1-1])<br />map[x1+2][y1-1] = map[x1][y1]+1;<br />if(x1-2 >= 0 && y1-1 >= 0 && !vis[x1-2][y1-1])<br />map[x1-2][y1-1] = map[x1][y1]+1;<br />}<br />void search(int x1,int y1,int x2,int y2)<br />{<br />int step = 1;<br />vis[x1][y1] = 1;<br />set(x1,y1);<br />//廣度優先搜尋過程<br />while(map[x2][y2] == 0)<br />{<br />for(int i = 0;i < 8;++i)<br />for(int j = 0;j < 8;++j)<br />if(map[i][j] == step)<br />{<br />vis[i][j] = 1;<br />set(i,j);<br />}<br />++step;<br />}<br />}<br />int main()<br />{<br />while(cin >> a >> b)<br />{<br />memset(map,0,sizeof(map));//初始化<br />memset(vis,0,sizeof(vis));//初始化<br />x1 = a[0] - 'a';<br />y1 = a[1] - '1';<br />x2 = b[0] - 'a';<br />y2 = b[1] - '1';<br />if(strcmp(a,b) == 0)<br />cout << "To get from "<< a <<" to "<< b << " takes 0 knight moves." <<endl;<br />else<br />{<br />search(x1,y1,x2,y2);<br />cout << "To get from "<< a <<" to "<< b << " takes "<< map[x2][y2] <<" knight moves." << endl;<br />}<br />}<br />return 0;<br />}