poj1279.Inlay Cutters（模拟 + 枚举）
Inlay Cutters
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2367 | Accepted: 995 |
Description
The factory cuts rectangular M × N granite plates into pieces using a special machine that is able to perform cuts in 4 different directions: vertically, horizontally, and diagonally at the angle of 45 degrees to the sides of the plate. Every cut is a straight line that starts and ends on the side of the plate. The factory has been ordered to produce tiles for the inlay, each tile of which is a 45 degrees right triangle. To reduce the time to deliver the tiles it was decided to take all triangles from the already cut plates. Information about all performed cuts is available and your task is to compute the number of triangles of any size that were produced.
Input
The input describes the cuts that were performed on a single rectangular plate. The first line of the input file contains three integer numbers M, N, and K, separated by spaces. M and N (1 ≤ M, N ≤ 50) are the dimensions of the plate, and K (0 ≤ K ≤ 296) is the number of cuts. Next K lines describe the cuts. ith cut is described by four integer numbers Xi,1, Yi,1, Xi,2, and Yi,2, separated by spaces, that represent the starting and ending point of the cut. Both starting (Xi,1, Yi,1) and ending (Xi,2, Yi,2) points of the cut are situated on the plate's border. Both points of the cut are different and the cut goes through the plate. Here, the coordinates by the X axis run from 0 to M, and the coordinates by the Y axis run from 0 to N. All cuts are different.
Output
Write to the output a single integer number - the number of triangles that were produced by the cuts.
Sample Input
7 4 6 6 0 7 1 1 4 1 0 0 4 4 0 0 0 4 4 0 2 7 2 7 0 3 4
Sample Output
8
Source
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 using namespace std; 5 int a[60 * 4][60 * 4] ; 6 bool map[60 * 4][60 * 4] ; 7 int col , row , cut ; 8 int x1 , y1 , x2 , y2 ; 9 int sum = 0 ; 10 int move[][4] = {{-1,0,0,1},{-1,1,1,1},{0,1,1,0},{1,1,1,-1},{1,0,0,-1},{1,-1,-1,-1},{0,-1,-1,0},{-1,-1,-1,1}}; 11 12 void hua1 () 13 { 14 for (int i = y1 ; i <= y2 ; i++) { 15 a[x1][i] += 1 ; 16 } 17 } 18 19 void hua2 () 20 { 21 for (int i = x1 ; i <= x2 ; i++) { 22 a[i][y1] += 1 ; 23 } 24 } 25 26 void hua3 () 27 { 28 int cnt = y2 - y1 + 1 ; 29 int x = x1 , y = y1 ; 30 while (cnt--) { 31 a[x][y] += 1 ; 32 x ++ , y ++ ; 33 } 34 } 35 36 void hua4 () 37 { 38 int cnt = x2 - x1 + 1 ; 39 int x = x1 , y = y1 ; 40 while (cnt--) { 41 a[x][y] += 1 ; 42 x ++ , y-- ; 43 } 44 } 45 46 bool judge1 (int x1 ,int y1 ,int x2 ,int y2) 47 { 48 int cnt = x2 - x1 - 1 ; 49 int x = x1 + 1 , y = y1 + 1 ; 50 while (cnt --) { 51 if (a[x][y] != 1 ) { 52 return false ; 53 } 54 x ++ , y ++ ; 55 } 56 return true ; 57 } 58 59 bool judge2 (int x1 , int y1 , int x2 , int y2) 60 { 61 int cnt = x2 - x1 - 1 ; 62 int x = x1 + 1 , y = y1 - 1 ; 63 while (cnt--) { 64 if (a[x][y] != 1) { 65 return false ; 66 } 67 x++ , y-- ; 68 } 69 return true ; 70 } 71 72 bool judge3 (int x1 , int y1 , int x2 , int y2) 73 { 74 int cnt = x2 - x1 - 1 ; 75 int x = x1 + 1 ; 76 while (cnt--) { 77 if (a[x][y1] != 1 ) { 78 return false ; 79 } 80 x ++ ; 81 } 82 return true ; 83 } 84 85 bool judge4 (int x1 , int y1 , int x2 , int y2) 86 { 87 int cnt = y2 - y1 - 1 ; 88 int y = y1 + 1 ; 89 while (cnt --) { 90 if (a[x1][y] != 1) { 91 return false ; 92 } 93 y ++ ; 94 } 95 return true ; 96 } 97 98 void bfs (int sx , int sy) 99 { 100 // printf ("sx = %d , sy = %d\n" , sx , sy) ; 101 for (int i = 0 ; i < 8 ; i++) { 102 int x1 = sx + move[i][0] , y1 = sy + move[i][1] ; 103 int x2 = sx + move[i][2] , y2 = sy + move[i][3] ; 104 if (x1 < 0 || y1 < 0 || x1 > row || y1 > col ) { 105 continue ; 106 } 107 if (x2 < 0 || y2 < 0 || x2 > row || y2 > col ) { 108 continue ; 109 } 110 if (a[x1][y1] * a[x2][y2] == 0 ) { 111 continue ; 112 } 113 while(a[x1][y1] * a[x2][y2] == 1 ) { 114 x1 += move[i][0] ; 115 y1 += move[i][1] ; 116 x2 += move[i][2] ; 117 y2 += move[i][3] ; 118 } 119 if (a[x1][y1] * a[x2][y2] == 0 ) { 120 continue ; 121 } 122 if (a[x1][y1] == 1 || a[x2][y2] == 1) { 123 continue ; 124 } 125 if (x1 != x2) { 126 int k = (y2 - y1) / (x2 - x1) ; 127 if (k == 1) { 128 if (x1 > x2) { 129 x1 ^= x2 ^= x1 ^= x2 ; 130 y1 ^= y2 ^= y1 ^= y2 ; 131 } 132 if (judge1 (x1 , y1 , x2 , y2) ) { 133 sum ++ ; 134 } 135 } 136 else if (k == -1) { 137 if (x1 > x2) { 138 x1 ^= x2 ^= x1 ^= x2 ; 139 y1 ^= y2 ^= y1 ^= y2 ; 140 } 141 if ( judge2 (x1 , y1 , x2 , y2) ) { 142 sum ++ ; 143 } 144 } 145 else { 146 if (x1 > x2) { 147 x1 ^= x2 ^= x1 ^= x2 ; 148 } 149 if ( judge3 (x1 , y1 , x2 , y2) ) { 150 sum ++ ; 151 } 152 } 153 } 154 else { 155 if (y1 > y2) { 156 y1 ^= y2 ^= y1 ^= y2 ; 157 } 158 if ( judge4 (x1 , y1 , x2 , y2) ) { 159 sum ++ ; 160 } 161 } 162 } 163 } 164 165 int main () 166 { 167 // freopen ("a.txt" , "r" , stdin ) ; 168 scanf ("%d%d%d" , &col , &row , &cut ) ; 169 col *= 4 , row *= 4 ; 170 for (int i = 0 ; i <= row ; i++) { 171 a[i][col] += 1 ; 172 a[i][0] += 1 ; 173 } 174 for (int i = 0 ; i <= col ; i++) { 175 a[row][i] += 1 ; 176 a[0][i] += 1 ; 177 } 178 while (cut--) { 179 scanf ("%d%d%d%d" , &y1 , &x1 , &y2 , &x2 ) ; 180 x1 *= 4 , y1 *= 4 , x2 *= 4 , y2 *= 4 ; 181 if (x1 == x2) { 182 if (y1 > y2) { 183 y1 ^= y2 ^= y1^= y2 ; 184 } 185 hua1 () ; 186 } 187 else if (y1 == y2) { 188 if (x1 > x2) { 189 x1 ^= x2 ^= x1 ^= x2 ; 190 } 191 hua2 () ; 192 } 193 else if ( (y2 - y1) / (x2 - x1) == 1) { 194 if (y1 > y2) { 195 y1 ^= y2 ^= y1^= y2 ; 196 x1 ^= x2 ^= x1 ^= x2 ; 197 } 198 hua3 () ; 199 } 200 else if ( (y2 - y1) / (x2 - x1) == -1 ) { 201 if (x1 > x2) { 202 y1 ^= y2 ^= y1^= y2 ; 203 x1 ^= x2 ^= x1 ^= x2 ; 204 } 205 hua4 () ; 206 } 207 } 208 209 /* for (int i = 0 ; i <= row ; i++) { 210 for (int j = 0 ; j<= col ; j++) { 211 printf ("%d " , a[i][j]) ; 212 } 213 puts ("") ; 214 }*/ 215 216 for (int i = 0 ; i <= row ; i++) { 217 for (int j = 0 ; j <= col ; j++) { 218 if (a[i][j] > 1) { 219 bfs (i , j) ; 220 } 221 } 222 } 223 printf ("%d\n" , sum ) ; 224 225 }
