The Donkey of Gui Zhou
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 308 Accepted Submission(s): 118
Problem Description
There was no donkey in the province of Gui Zhou, China. A trouble maker shipped one and put it in the forest which could be considered as an N×N grid. The coordinates of the up-left cell is (0,0) , the down-right cell is (N-1,N-1) and the cell below the up-left cell is (1,0)..... A 4×4 grid is shown below:
Input
There are several test cases.
In each test case: First line is an integer N, meaning that the forest is a N×N grid.
The second line contains three integers R, C and D, meaning that the donkey is in the cell (R,C) when they started running, and it's original direction is D. D can be 0, 1, 2 or 3. 0 means east, 1 means south , 2 means west, and 3 means north.
The third line has the same format and meaning as the second line, but it is for the tiger.
The input ends with N = 0. ( 2 <= N <= 1000, 0 <= R, C < N)
In each test case: First line is an integer N, meaning that the forest is a N×N grid.
The second line contains three integers R, C and D, meaning that the donkey is in the cell (R,C) when they started running, and it's original direction is D. D can be 0, 1, 2 or 3. 0 means east, 1 means south , 2 means west, and 3 means north.
The third line has the same format and meaning as the second line, but it is for the tiger.
The input ends with N = 0. ( 2 <= N <= 1000, 0 <= R, C < N)
Output
For each test case, if the donkey and the tiger would meet in a cell, print the coordinate of the cell where they meet first time. If they would never meet, print -1 instead.
Sample Input
2
0 0 0
0 1 2
4
0 1 0
3 2 0
0
Sample Output
-1
1 3
Source
杭州网络预选赛第三题，很简单的一道题，只要按照给定的走法一步上步模拟即可
代码是比赛时候写的，略难看。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define UP 3 5 #define DOWN 1 6 #define LEFT 2 7 #define RIGHT 0 8 9 using namespace std; 10 11 class POINT 12 { 13 public: 14 int x; 15 int y; 16 bool operator != (const POINT &p) const 17 { 18 if(x==p.x&&y==p.y) 19 return false; 20 return true; 21 } 22 }; 23 24 int n; 25 26 27 int donkey_dir,tiger_dir; 28 bool donkey_end,tiger_end; 29 bool donkey_map[1010][1010],tiger_map[1010][1010]; 30 POINT donkey_current,tiger_current; 31 32 int main() 33 { 34 while(scanf("%d",&n)) 35 { 36 if(n==0) 37 break; 38 memset(donkey_map,false,sizeof(donkey_map)); 39 memset(tiger_map,false,sizeof(tiger_map)); 40 donkey_end=tiger_end=false; 41 int r,c,d; 42 scanf("%d %d %d",&r,&c,&d); 43 POINT temp; 44 temp.x=r+1; 45 temp.y=c+1; 46 donkey_current=temp; 47 donkey_dir=d; 48 scanf("%d %d %d",&r,&c,&d); 49 temp.x=r+1; 50 temp.y=c+1; 51 tiger_current=temp; 52 tiger_dir=d; 53 for(int i=0;i<=n+1;i++) 54 { 55 tiger_map[i][0]=tiger_map[i][n+1]=donkey_map[i][0]=donkey_map[i][n+1]=true; 56 tiger_map[0][i]=tiger_map[n+1][i]=donkey_map[0][i]=donkey_map[n+1][i]=true; 57 } 58 59 while(tiger_current!=donkey_current) 60 { 61 if(donkey_end&&tiger_end) 62 break; 63 if(!donkey_end) 64 { 65 int x=donkey_current.x; 66 int y=donkey_current.y; 67 if(donkey_dir==UP) 68 { 69 if(donkey_map[x-1][y]) 70 { 71 if(donkey_map[x][y+1]) 72 donkey_end=true; 73 else 74 { 75 temp.x=x; 76 temp.y=y+1; 77 donkey_current=temp; 78 donkey_map[x][y]=true; 79 donkey_dir=RIGHT; 80 } 81 } 82 else 83 { 84 temp.x=x-1; 85 temp.y=y; 86 donkey_current=temp; 87 donkey_map[x][y]=true; 88 } 89 } 90 else if(donkey_dir==RIGHT) 91 { 92 if(donkey_map[x][y+1]) 93 { 94 if(donkey_map[x+1][y]) 95 donkey_end=true; 96 else 97 { 98 temp.x=x+1; 99 temp.y=y; 100 donkey_current=temp; 101 donkey_map[x][y]=true; 102 donkey_dir=DOWN; 103 } 104 } 105 else 106 { 107 temp.x=x; 108 temp.y=y+1; 109 donkey_current=temp; 110 donkey_map[x][y]=true; 111 } 112 } 113 else if(donkey_dir==DOWN) 114 { 115 if(donkey_map[x+1][y]) 116 { 117 if(donkey_map[x][y-1]) 118 donkey_end=true; 119 else 120 { 121 temp.x=x; 122 temp.y=y-1; 123 donkey_current=temp; 124 donkey_map[x][y]=true; 125 donkey_dir=LEFT; 126 } 127 } 128 else 129 { 130 temp.x=x+1; 131 temp.y=y; 132 donkey_current=temp; 133 donkey_map[x][y]=true; 134 } 135 } 136 else if(donkey_dir==LEFT) 137 { 138 if(donkey_map[x][y-1]) 139 { 140 if(donkey_map[x-1][y]) 141 donkey_end=true; 142 else 143 { 144 temp.x=x-1; 145 temp.y=y; 146 donkey_current=temp; 147 donkey_map[x][y]=true; 148 donkey_dir=UP; 149 } 150 } 151 else 152 { 153 temp.x=x; 154 temp.y=y-1; 155 donkey_current=temp; 156 donkey_map[x][y]=true; 157 } 158 } 159 } 160 if(!tiger_end) 161 { 162 int x=tiger_current.x; 163 int y=tiger_current.y; 164 if(tiger_dir==UP) 165 { 166 if(tiger_map[x-1][y]) 167 { 168 if(tiger_map[x][y-1]) 169 tiger_end=true; 170 else 171 { 172 temp.x=x; 173 temp.y=y-1; 174 tiger_current=temp; 175 tiger_map[x][y]=true; 176 tiger_dir=LEFT; 177 } 178 } 179 else 180 { 181 temp.x=x-1; 182 temp.y=y; 183 tiger_current=temp; 184 tiger_map[x][y]=true; 185 } 186 } 187 else if(tiger_dir==RIGHT) 188 { 189 if(tiger_map[x][y+1]) 190 { 191 if(tiger_map[x-1][y]) 192 tiger_end=true; 193 else 194 { 195 temp.x=x-1; 196 temp.y=y; 197 tiger_current=temp; 198 tiger_map[x][y]=true; 199 tiger_dir=UP; 200 } 201 } 202 else 203 { 204 temp.x=x; 205 temp.y=y+1; 206 tiger_current=temp; 207 tiger_map[x][y]=true; 208 } 209 } 210 else if(tiger_dir==DOWN) 211 { 212 if(tiger_map[x+1][y]) 213 { 214 if(tiger_map[x][y+1]) 215 tiger_end=true; 216 else 217 { 218 temp.x=x; 219 temp.y=y+1; 220 tiger_current=temp; 221 tiger_map[x][y]=true; 222 tiger_dir=RIGHT; 223 } 224 } 225 else 226 { 227 temp.x=x+1; 228 temp.y=y; 229 tiger_current=temp; 230 tiger_map[x][y]=true; 231 } 232 } 233 else if(tiger_dir==LEFT) 234 { 235 if(tiger_map[x][y-1]) 236 { 237 if(tiger_map[x+1][y]) 238 tiger_end=true; 239 else 240 { 241 temp.x=x+1; 242 temp.y=y; 243 tiger_current=temp; 244 tiger_map[x][y]=true; 245 tiger_dir=DOWN; 246 } 247 } 248 else 249 { 250 temp.x=x; 251 temp.y=y-1; 252 tiger_current=temp; 253 tiger_map[x][y]=true; 254 } 255 } 256 } 257 } 258 if(tiger_current!=donkey_current) 259 cout<<"-1"<<endl; 260 else 261 cout<<tiger_current.x-1<<" "<<tiger_current.y-1<<endl; 262 } 263 264 return 0; 265 }