USACO 5.1 Starry Night(模拟题)
把所有的连通块，左上角，右下角的坐标当作一个矩形存起来。然后枚举，跟之前的是不是相同。
注意：连通块转化成矩阵的时候，只有一起连通的的点，在小的矩形里才为1，而不是str[i][j] =='1'，这里错了次。。写的很麻烦。。
1 /* 2 ID:cuizhe 3 LANG: C++ 4 TASK: starry 5 */ 6 #include <cstdio> 7 #include <cstring> 8 #include <iostream> 9 using namespace std; 10 struct node 11 { 12 int n,m; 13 int lx,ly,rx,ry; 14 int flag; 15 } p[2001]; 16 char str[101][101]; 17 int o[101][101],w,h; 18 int t1[101][101],t2[101][101]; 19 int lx,ly,rx,ry; 20 int kk[2001]; 21 int a[8] = {1,1,1,0,0,-1,-1,-1}; 22 int b[8] = {0,-1,1,-1,1,0,1,-1}; 23 void CL() 24 { 25 lx = 1000000; 26 ly = 1000000; 27 rx = -1; 28 ry = -1; 29 } 30 void dfs(int x,int y,int num) 31 { 32 int i; 33 lx = min(x,lx); 34 ly = min(y,ly); 35 rx = max(rx,x); 36 ry = max(ry,y); 37 for(i = 0; i < 8; i ++) 38 { 39 if(x+a[i]>=0&&x+a[i] < h&&y+b[i] >= 0&&y+b[i] < w&&!o[x+a[i]][y+b[i]]&&str[x+a[i]][y+b[i]] == '1') 40 { 41 o[x+a[i]][y+b[i]] = num; 42 dfs(x+a[i],y+b[i],num); 43 } 44 } 45 return ; 46 } 47 int fun(int x,int y) 48 { 49 int i,j,z,n,m; 50 n = p[x].n; 51 m = p[x].m; 52 if(p[x].n == p[y].n&&p[x].m == p[y].m) 53 { 54 z = 1; 55 for(i = 0; i < n&&z; i ++) //相等 56 { 57 for(j = 0; j < m&&z; j ++) 58 { 59 if(t1[i][j] != t2[i][j]) 60 z = 0; 61 } 62 } 63 if(z) return 1; 64 z = 1; 65 for(i = 0; i < n&&z; i ++) //旋转180 66 { 67 for(j = 0; j < m&&z; j ++) 68 { 69 if(t1[i][j] != t2[n-i-1][m-j-1]) 70 z = 0; 71 } 72 } 73 if(z) return 1; 74 z = 1; 75 for(i = 0; i < n&&z; i ++) //翻转 76 { 77 for(j = 0; j < m&&z; j ++) 78 { 79 if(t1[i][j] != t2[i][m-j-1]) 80 z = 0; 81 } 82 } 83 if(z) return 1; 84 z = 1; 85 for(i = 0; i < n&&z; i ++) //翻转然后旋转180 86 { 87 for(j = 0; j < m&&z; j ++) 88 { 89 if(t1[i][j] != t2[n-i-1][j]) 90 z = 0; 91 } 92 } 93 if(z) return 1; 94 } 95 if(p[x].n == p[y].m&&p[x].m == p[y].n) 96 { 97 z = 1; 98 for(i = 0; i < n&&z; i ++) //旋转90 99 { 100 for(j = 0; j < m&&z; j ++) 101 { 102 if(t1[i][j] != t2[j][n-i-1]) 103 z = 0; 104 } 105 } 106 if(z) return 1; 107 z = 1; 108 for(i = 0; i < n&&z; i ++) //旋转270 109 { 110 for(j = 0; j < m&&z; j ++) 111 { 112 if(t1[i][j] != t2[m-j-1][i]) 113 z = 0; 114 } 115 } 116 if(z) return 1; 117 z = 1; 118 for(i = 0; i < n&&z; i ++) //先翻转后旋转90 119 { 120 for(j = 0; j < m&&z; j ++) 121 { 122 if(t1[i][j] != t2[m-j-1][n-i-1]) 123 z = 0; 124 } 125 } 126 if(z) return 1; 127 z = 1; 128 for(i = 0; i < n&&z; i ++) //先翻转后旋转270 129 { 130 for(j = 0; j < m&&z; j ++) 131 { 132 if(t1[i][j] != t2[j][i]) 133 z = 0; 134 } 135 } 136 if(z) return 1; 137 } 138 return 0; 139 } 140 int judge(int x,int y) 141 { 142 int i,j; 143 if(p[x].n == p[y].n&&p[x].m == p[y].m) 144 ; 145 else if(p[x].n == p[y].m&&p[x].m == p[y].n) 146 ; 147 else 148 return 0; 149 for(i = p[x].lx; i <= p[x].rx; i ++)//转化为两个01矩阵 150 { 151 for(j = p[x].ly; j <= p[x].ry; j ++) 152 { 153 if(o[i][j] == x) 154 t1[i-p[x].lx][j-p[x].ly] = 1; 155 else 156 t1[i-p[x].lx][j-p[x].ly] = 0; 157 } 158 } 159 for(i = p[y].lx; i <= p[y].rx; i ++) 160 { 161 for(j = p[y].ly; j <= p[y].ry; j ++) 162 { 163 if(o[i][j] == y) 164 t2[i-p[y].lx][j-p[y].ly] = 1; 165 else 166 t2[i-p[y].lx][j-p[y].ly] = 0; 167 } 168 } 169 170 if(fun(x,y)) 171 return 1; 172 else 173 return 0; 174 } 175 int main() 176 { 177 int i,j,num; 178 freopen("starry.in","r",stdin); 179 freopen("starry.out","w",stdout); 180 scanf("%d%d",&w,&h); 181 for(i = 0; i < h; i ++) 182 { 183 scanf("%s",str[i]); 184 } 185 num = 1; 186 for(i = 0; i < h; i ++) 187 { 188 for(j = 0; j < w; j ++) 189 { 190 if(!o[i][j]&&str[i][j] == '1') 191 { 192 o[i][j] = num; 193 CL(); 194 dfs(i,j,num); 195 p[num].lx = lx; 196 p[num].ly = ly; 197 p[num].rx = rx; 198 p[num].ry = ry; 199 p[num].flag = 0; 200 p[num].n = rx - lx + 1; 201 p[num].m = ry - ly + 1; 202 num ++; 203 } 204 } 205 } 206 for(i = 2; i < num; i ++) 207 { 208 for(j = 1; j < i; j ++) 209 { 210 if(p[j].flag) continue; 211 if(judge(i,j)) 212 { 213 p[i].flag = j; 214 break; 215 } 216 } 217 } 218 num = 1; 219 for(i = 0; i < h; i ++) 220 { 221 for(j = 0; j < w; j ++) 222 { 223 if(str[i][j] == '1') 224 { 225 if(p[o[i][j]].flag) 226 { 227 kk[o[i][j]] = kk[p[o[i][j]].flag]; 228 } 229 else 230 { 231 if(!kk[o[i][j]]) 232 kk[o[i][j]] = num ++; 233 } 234 printf("%c",kk[o[i][j]]-1+'a'); 235 } 236 else 237 printf("0"); 238 } 239 printf("\n"); 240 } 241 return 0; 242 }
