枚举 POJ 2965 The Pilots Brothers' refrigerator

 

题目地址：http://poj.org/problem?id=2965

 1 /*  2  题意：4*4的矩形，改变任意点，把所有'+'变成'-',，每一次同行同列的都会反转，求最小步数，并打印方案  3  4  DFS：把'+'记为1， '-'记为0  5  1. 从(1, 1)走到(4, 4)，每一点DFS两次(改点反转或不反转)；used记录反转的位置  6  详细解释：http://poj.org/showmessage?message_id=346723  7  2. 比较巧妙的解法：抓住'+'位置反转，'-'位置相对不反转的特点，从状态上考虑  8  详细解释：http://poj.org/showmessage?message_id=156561  9  3. 枚举步骤数(1~16)，暴力解法，耗时大  10  详细解释：http://poj.org/showmessage?message_id=343281  11  4. 网上还有其他解法：高斯消元，BFS，+位运算等等  12  13  注意：反转时十字形中心位置多反转了两次，要再反转一次  14  15  我还是DFS写不出来。。。  16 */  17 #include <cstdio>  18 #include <iostream>  19 #include <algorithm>  20 #include <cstring>  21 #include <cmath>  22 #include <string>  23 #include <map>  24 #include <queue>  25 #include <vector>  26 using namespace std;  27  28 const int MAXN = 1e6 + 10;  29 const int INF = 0x3f3f3f3f;  30 int a[5][5];  31 int used[5][5];  32  33 bool ok(void)  34 {  35 for (int i=1; i<=4; ++i)  36  {  37 for (int j=1; j<=4; ++j)  38 if (a[i][j] == 1) return false;  39  }  40  41 return true;  42 }  43  44 void change(int x, int y)  45 {  46 for (int i=1; i<=4; ++i)  47  {  48 a[x][i] = a[x][i] ? 0 : 1;  49 a[i][y] = a[i][y] ? 0 : 1;  50  }  51 a[x][y] = a[x][y] ? 0 : 1;  52 used[x][y] = used[x][y] ? 0 : 1;  53 }  54  55 bool DFS(int x, int y)  56 {  57 if (x == 4 && y == 4)  58  {  59 if (ok ()) return true;  60  61  change (x, y);  62 if (ok ()) return true;  63  64  change (x, y);  65 return false;  66  }  67 int nx, ny;  68 if (y == 4) nx = x + 1, ny = 1;  69 else nx = x, ny = y + 1;  70  71 if (DFS (nx, ny)) return true;  72  73  change (x, y);  74 if (DFS (nx, ny)) return true;  75  76  change (x, y);  77 return false;  78 }  79  80 void work(void)  81 {  82 if (DFS (1, 1))  83  {  84 int ans = 0;  85 for (int i=1; i<=4; ++i)  86  {  87 for (int j=1; j<=4; ++j)  88  {  89 if (used[i][j]) ans++;  90  }  91  }  92 printf ("%d\n", ans);  93 for (int i=1; i<=4; ++i)  94 for (int j=1; j<=4; ++j)  95 if (used[i][j]) printf ("%d %d\n", i, j);  96  }  97 else printf ("WA\n");  98 }  99 100 int main(void) //POJ 2965 The Pilots Brothers' refrigerator 101 { 102 //freopen ("B.in", "r", stdin); 103 104 char ch; 105 for (int i=1; i<=4; ++i) 106  { 107 for (int j=1; j<=4; ++j) 108  { 109 scanf ("%c", &ch); 110 a[i][j] = (ch == '-') ? 0 : 1; 111 //printf ("%d ", a[i][j]); 112  } 113 getchar (); //puts (""); 114  } 115 116  work (); 117 118 return 0; 119 } 120 121 /* 122 #include <cstdio> 123 #include <iostream> 124 #include <algorithm> 125 #include <cstring> 126 #include <cmath> 127 #include <string> 128 #include <map> 129 #include <queue> 130 #include <vector> 131 using namespace std; 132 133 const int MAXN = 1e6 + 10; 134 const int INF = 0x3f3f3f3f; 135 bool con[5][5]; 136 int a[5][5]; 137 struct NODE 138 { 139  int x, y; 140 }node[MAXN]; 141 142 void work(void) 143 { 144  for (int i=1; i<=4; ++i) 145  { 146  for (int j=1; j<=4; ++j) 147  { 148  if (a[i][j] == 1) 149  { 150  con[i][j] = !con[i][j]; 151  for (int k=1; k<=4; ++k) 152  { 153  con[i][k] = !con[i][k]; 154  con[k][j] = !con[k][j]; 155  } 156  } 157  } 158  } 159  int ans = 0; 160  for (int i=1; i<=4; ++i) 161  { 162  for (int j=1; j<=4; ++j) 163  { 164  if (con[i][j] == true) 165  { 166  ans++; node[ans].x = i; node[ans].y = j; 167  } 168  } 169  } 170  printf ("%d\n", ans); 171  for (int i=1; i<=ans; ++i) 172  printf ("%d %d\n", node[i].x, node[i].y); 173 } 174 175 int main(void) //POJ 2965 The Pilots Brothers' refrigerator 176 { 177  //freopen ("B.in", "r", stdin); 178 179  char ch; 180  for (int i=1; i<=4; ++i) 181  { 182  for (int j=1; j<=4; ++j) 183  { 184  scanf ("%c", &ch); 185  a[i][j] = (ch == '-') ? 0 : 1; 186  } 187  getchar (); 188  } 189 190  work (); 191 192  return 0; 193 } 194 */ 195 196 /* 197 #include <cstdio> 198 #include <iostream> 199 #include <algorithm> 200 #include <cstring> 201 #include <cmath> 202 #include <string> 203 #include <map> 204 #include <queue> 205 #include <vector> 206 using namespace std; 207 208 const int MAXN = 1e6 + 10; 209 const int INF = 0x3f3f3f3f; 210 int a[5][5]; 211 struct NODE 212 { 213  int x, y; 214 }node[MAXN]; 215 int k; 216 bool flag; 217 218 bool ok(void) 219 { 220  for (int i=1; i<=4; ++i) 221  { 222  for (int j=1; j<=4; ++j) 223  if (a[i][j] == 1) return false; 224  } 225 226  return true; 227 } 228 229 void change(int x, int y) 230 { 231  for (int i=1; i<=4; ++i) 232  { 233  a[x][i] = !a[x][i]; 234  a[i][y] = !a[i][y]; 235  } 236  a[x][y] = !a[x][y]; 237 } 238 239 void DFS(int x, int y, int num, int cnt, int kk) 240 { 241  if (num == cnt) 242  { 243  flag = ok (); 244  k = kk; 245  return ; 246  } 247  for (int i=x; i<=4; ++i) 248  { 249  int j; 250  if (i == x) j = y + 1; 251  else j = 1; 252  for (; j<=4; ++j) 253  { 254  node[kk].x = i; 255  node[kk].y = j; 256  change (i, j); 257  DFS (i, j, num+1, cnt, kk+1); 258  if (flag) return ; 259  change (i, j); 260  } 261  } 262 } 263 264 void work(void) 265 { 266  int cnt; 267  for (cnt=1; cnt<=16; ++cnt) 268  { 269  flag = false; k = 0; 270  DFS (1, 0, 0, cnt, 1); 271  if (flag) break; 272  } 273 274  printf ("%d\n", cnt); 275  for (int i=1; i<=k - 1; ++i) 276  printf ("%d %d\n", node[i].x, node[i].y); 277 } 278 279 int main(void) //POJ 2965 The Pilots Brothers' refrigerator 280 { 281  //freopen ("B.in", "r", stdin); 282 283  char ch; 284  for (int i=1; i<=4; ++i) 285  { 286  for (int j=1; j<=4; ++j) 287  { 288  scanf ("%c", &ch); 289  a[i][j] = (ch == '-') ? 0 : 1; 290  } 291  getchar (); 292  } 293 294  work (); 295 296  return 0; 297 } 298 */