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 }