poj2488 A Knight's Journey 简单DFS 注意搜索步骤
这一题是比较简单的搜索题，但是题目要求走法是按照字典序的，这样要求搜索时注意顺序。
1 ///2014.3.10 2 ///poj2488 3 4 //16MS 5 6 /** 7 *简单的DFS，但是题目有个坑人的地方，就是字典序 8 *因为要求字典序，在搜索时要注意搜索步骤 9 *在搜索步骤上浪费了两个小时啊！！！ 10 */ 11 12 #include <iostream> 13 #include <cstdio> 14 using namespace std; 15 16 int p,q; 17 int coordinate_x[30]; ///记录每一步的坐标 18 int coordinate_y[30]; 19 bool findit; ///标记有没有找到走法 20 bool step[30][30]; ///标记每一个位置有没有走过 21 22 void dfs(int x,int y,int deep) 23 { 24 if( findit ) 25 return; 26 if( deep==p*q ){ 27 findit = true; 28 coordinate_x[deep] = x; 29 coordinate_y[deep] = y; 30 return; 31 } 32 33 step[x][y] = true; 34 coordinate_x[deep] = x; 35 coordinate_y[deep] = y; 36 if( x-1>=1 && x-1<=p && y-2>=1 && y-2<=q && !step[x-1][y-2] ){ 37 dfs(x-1,y-2,deep+1); 38 step[x-1][y-2] = false; 39 } 40 if( x+1>=1 && x+1<=p && y-2>=1 && y-2<=q && !step[x+1][y-2] ){ 41 dfs(x+1,y-2,deep+1); 42 step[x+1][y-2] = false; 43 } 44 if( x-2>=1 && x-2<=p && y-1>=1 && y-1<=q && !step[x-2][y-1] ){ 45 dfs(x-2,y-1,deep+1); 46 step[x-2][y-1] = false; 47 } 48 if( x+2>=1 && x+2<=p && y-1>=1 && y-1<=q && !step[x+2][y-1] ){ 49 dfs(x+2,y-1,deep+1); 50 step[x+2][y-1] = false; 51 } 52 if( x-2>=1 && x-2<=p && y+1>=1 && y+1<=q && !step[x-2][y+1] ){ 53 dfs(x-2,y+1,deep+1); 54 step[x-2][y+1] = false; 55 } 56 if( x+2>=1 && x+2<=p && y+1>=1 && y+1<=q && !step[x+2][y+1] ){ 57 dfs(x+2,y+1,deep+1); 58 step[x+2][y+1] = false; 59 } 60 if( x-1>=1 && x-1<=p && y+2>=1 && y+2<=q && !step[x-1][y+2] ){ 61 dfs(x-1,y+2,deep+1); 62 step[x-1][y+2] = false; 63 } 64 if( x+1>=1 && x+1<=p && y+2>=1 && y+2<=q && !step[x+1][y+2] ){ 65 dfs(x+1,y+2,deep+1); 66 step[x+1][y+2] = false; 67 } 68 } 69 70 int main() 71 { 72 // freopen("in","r",stdin); 73 // freopen("out","w",stdout); 74 75 int n; 76 int cas = 1; 77 scanf("%d",&n); 78 while( n-- ){ 79 for(int i=0 ; i<27 ; i++) 80 for(int j=0 ; j<27 ; j++) 81 step[i][j] = false; 82 findit = false; 83 scanf("%d%d",&p,&q); 84 dfs(1,1,1); 85 if( findit ){ 86 printf("Scenario #%d:\n",cas); 87 for(int i=1 ; i<=p*q ; i++) 88 printf("%c%d",'A'+coordinate_y[i]-1,coordinate_x[i]); 89 printf("\n\n"); 90 } 91 else 92 printf("Scenario #%d:\nimpossible\n\n",cas); 93 cas++; 94 } 95 return 0; 96 }
在discuss中看到一个打表的代码，我怎么没想到呢~贴下来供以后启发思维
1 #include <iostream> 2 using namespace std; 3 4 5 int main() 6 { 7 freopen("in.txt","r",stdin); 8 int i,n,p,q; 9 cin>>n; 10 for(i=1;i<=n;i++) 11 { 12 cin>>p>>q; 13 cout<<"Scenario #"<<i<<":"<<endl; 14 if(p==1 &&q ==1) 15 cout<<"A1"<<endl; 16 else if(p==3 && q==4) 17 cout<<"A1C2A3B1D2B3C1A2C3D1B2D3"<<endl; 18 else if(p==3 && q==7) 19 cout<<"A1B3D2F1G3E2G1F3E1G2E3C2A3B1C3A2C1D3B2D1F2"<<endl; 20 else if(p==3 && q==8) 21 cout<<"A1B3C1A2C3D1B2D3E1G2E3C2A3B1D2F1H2F3G1E2G3H1F2H3"<<endl; 22 else if(p==4 && q==3) 23 cout<<"A1B3C1A2B4C2A3B1C3A4B2C4"<<endl; 24 else if(p==4 && q==5) 25 cout<<"A1B3C1A2B4D3E1C2D4E2C3A4B2D1E3C4A3B1D2E4"<<endl; 26 else if(p==4 && q==6) 27 cout<<"A1B3C1A2B4C2D4E2F4D3E1F3D2B1A3C4B2A4C3E4F2D1E3F1"<<endl; 28 else if(p==5 && q==4) 29 cout<<"A1B3A5C4D2B1A3B5D4C2B4A2C1D3C5A4B2D1C3D5"<<endl; 30 else if(p==5 && q==5) 31 cout<<"A1B3A5C4A3B1D2E4C5A4B2D1C3B5D4E2C1A2B4D5E3C2E1D3E5"<<endl; 32 else if(p==6 && q==4) 33 cout<<"A1B3A5C6D4B5D6C4D2B1A3C2B4A2C1D3B2D1C3D5B6A4C5A6"<<endl; 34 else if(p==7 && q==3) 35 cout<<"A1B3C1A2C3B1A3C2B4A6C7B5A7C6A5B7C5A4B2C4B6"<<endl; 36 else if(p==7 && q==4) 37 cout<<"A1B3A5B7D6B5A7C6D4C2A3B1D2C4B2A4B6D7C5A6C7D5B4D3C1A2C3D1"<<endl; 38 else if(p==8 && q==3) 39 cout<<"A1B3C1A2B4C2A3B1C3A4B2C4A5B7C5A6B8C6A7B5C7A8B6C8"<<endl; 40 else 41 cout<<"impossible"<<endl; 42 cout<<endl; 43 } 44 return 0; 45 46 }
