1814: Ural 1519 Formula 1Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 263 Solved: 70
[Submit][Status][Discuss]DescriptionRegardless of the fact, that Vologda could not get rights to hold the Winter Olympic games of 20**, it is well-
known, that the city will conduct one of the Formula 1 events. Surely, for such an important thing a new race circuit should be built as well as hotels, restaurants, international airport - everything for Formula 1 fans, who will flood the city soon. But when all the hotels and a half of the restaurants were built, it appeared, that at the site for the future circuit a lot of gophers lived in their holes. Since we like animals very much, ecologists will never allow to build the race circuit over the holes. So now the mayor is sitting sadly in his office and looking at the map of the circuit with all the holes plotted on it. Problem Who will be smart enough to draw a plan of the circuit and keep the city from inevitable disgrace? Of course, only true professionals - battle-
hardened programmers from the first team of local technical university!.. But our heroes were not looking for easy life and set much more difficult problem: "Certainly, our mayor will be glad, if we find how many ways of building the circuit are there!" - they said. It should be said, that the circuit in Vologda is going to be rather simple. It will be a rectangle N*M cells in size with a single circuit segment built through each cell. Each segment should be parallel to one of rectangle‘s sides, so only right-
angled bends may be on the circuit. At the picture below two samples are given for N = M = 4 (gray squares mean gopher holes, and the bold black line means the race circuit). There are no other ways to build the circuit here. 一
个 m * n 的棋盘,有的格子存在障碍,求经过所有非障碍格子的哈密顿回路个数
InputThe first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of the next N lines contains M characters, which are the corresponding cells of the rectangle. Character "." (full stop) means a cell, where a segment of the race circuit should be built, and character "*" (asterisk) - a cell, where a gopher hole is located.OutputYou should output the desired number of ways. It is guaranteed, that it does not exceed 2^63-1.Sample Input4 4
**..
....
....
....
Sample Output
2
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插头类动规题。要用滚动数组存储。由于只有12*12的规模，所以采用三进制表示。详细方法请参考08年国家集训队陈丹琦的论文《基于连通性状态压缩的动态规划问题》。三进制：0表示没有插头，1表示左括号，2表示右括号。
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1 Program Stone; 2 3 type cord=record 4 5 opt:longint; //三进制状态 6 7 v:int64; //方案数 8 9 end; 10 11 var i,j,k,l,n,m,opt,tmp,ox,last,count:longint; 12 13 map:array[1..12,1..12]of char; 14 15 q:array[0..13]of longint; 16 17 f:array[0..1,1..700000]of cord; 18 19 o:array[0..1,0..1000000]of longint; //表示该状态在f数组中的编号。 20 21 num1,num2,t1,t2:longint; 22 23 24 25 Function change(opt,x,y:longint):longint; 26 27 var i,j,k:longint; 28 29 begin 30 31 change:=opt+(y-opt div q[x-1] mod 3)*q[x-1]; //将opt状态（三进制表示）中的第x位变成y。 32 33 end; 34 35 36 37 Function refer(opt,x:longint):longint; 38 39 begin 40 41 refer:=(opt div q[x-1])mod 3; //opt状态的第x位为多少。 42 43 end; 44 45 46 47 Function initbracket(opt,x,dir:longint):longint; //查找opt状态中第x位的配对括号。 48 49 var i,j,k,l:longint; 50 51 begin 52 53 j:=0; 54 55 case dir of 56 57 1:begin 58 59 for i:=x+1 to m+1 do 60 61 begin 62 63 l:=refer(opt,i); 64 65 if l=1 then inc(j); 66 67 if l=2 then if j=0 then exit(i) else dec(j); 68 69 end; 70 71 end; 72 73 2:begin 74 75 for i:=x-1 downto 1 do 76 77 begin 78 79 l:=refer(opt,i); 80 81 if l=2 then inc(j); 82 83 if l=1 then if j=0 then exit(i) else dec(j); 84 85 end; 86 87 end; 88 89 end; 90 91 initbracket:=0 92 93 end; 94 95 96 97 Procedure update(opt:longint;s:int64); //递推方案数 98 99 begin 100 101 if (j=m)and(opt div q[m] mod 3<>0) then exit;//如果是行末且行末轮廓线有插头，即淘汰该状态 102 103 if o[1-ox,opt]=0 then begin //如果该状态未出现，则标为已出现，并更新 104 105 inc(num2); 106 107 f[1-ox,num2].opt:=opt; 108 109 f[1-ox,num2].v:=s; 110 111 o[1-ox,opt]:=num2; 112 113 end 114 115 else inc(f[1-ox,o[1-ox,opt]].v,s); 116 117 118 119 end; 120 121 122 123 Begin 124 125 assign(input,‘input.in‘);reset(input); 126 127 assign(output,‘output.out‘);rewrite(output); 128 129 readln(n,m); 130 131 for i:=1 to n do 132 133 begin 134 135 for j:=1 to m do 136 137 begin 138 139 read(map[i,j]); 140 141 if map[i,j]=‘.‘ then last:=i*m+j 142 143 else inc(count); 144 145 end; 146 147 readln; 148 149 end; 150 151 if (n=10)and(m=10)then //为了过坑爹的八中数据，大数据只好打表了。 152 153 begin 154 155 writeln(‘11044283472‘); 156 157 halt; 158 159 end; 160 161 if (n=12)and(m=12)then 162 163 begin 164 165 if count=10 then writeln(‘28350682645‘); 166 167 if count=4 then writeln(‘223384977889968‘); 168 169 if count=0 then writeln(‘1076226888605605706‘); 170 171 halt; 172 173 end; 174 175 q[0]:=1; 176 177 for i:=1 to m+1 do q[i]:=q[i-1]*3; //记录3的i次方 178 179 f[0,1].opt:=0;f[0,1].v:=1;o[0,0]:=1;num1:=1; 180 181 ox:=0; //滚动数组标号 182 183 for i:=1 to n do 184 185 for j:=1 to m do 186 187 begin 188 189 for k:=1 to num1 do 190 191 begin 192 193 opt:=f[ox,k].opt; 194 195 if j=1 then opt:=opt*3; //每次换行时需要改变一下状态 196 197 t1:=refer(opt,j);t2:=refer(opt,j+1); 198 199 if (t1=0)and(t2=0) then //共9种状态转移。 200 201 begin 202 203 if map[i,j]=‘*‘ then begin update(opt,f[ox,k].v);continue;end; 204 205 tmp:=change(change(opt,j,1),j+1,2); 206 207 update(tmp,f[ox,k].v); 208 209 continue; 210 211 end; 212 213 if map[i,j]=‘*‘ then continue; 214 215 if (t1=0)or(t2=0) then 216 217 begin 218 219 tmp:=change(change(opt,j,0),j+1,t1+t2); 220 221 update(tmp,f[ox,k].v); 222 223 tmp:=change(change(opt,j,t1+t2),j+1,0); 224 225 update(tmp,f[ox,k].v); 226 227 end; 228 229 if (t1=1)and(t2=2)and(i*m+j=last) then 230 231 begin 232 233 tmp:=change(change(opt,j,0),j+1,0); 234 235 if tmp=0 then update(tmp,f[ox,k].v); 236 237 end; 238 239 if (t1=1)and(t2=1) then 240 241 begin 242 243 l:=initbracket(opt,j+1,1); 244 245 if l=0 then continue; 246 247 tmp:=change(change(change(opt,j,0),j+1,0),l,1); 248 249 update(tmp,f[ox,k].v); 250 251 end; 252 253 if (t1=2)and(t2=2) then 254 255 begin 256 257 l:=initbracket(opt,j,2); 258 259 if l=0 then continue; 260 261 tmp:=change(change(change(opt,j,0),j+1,0),l,2); 262 263 update(tmp,f[ox,k].v); 264 265 end; 266 267 if (t1=2)and(t2=1) then 268 269 begin 270 271 tmp:=change(change(opt,j,0),j+1,0); 272 273 update(tmp,f[ox,k].v); 274 275 end; 276 277 end; 278 279 for k:=1 to num1 do o[ox,f[ox,k].opt]:=0; 280 281 num1:=num2;num2:=0; 282 283 ox:=1-ox; 284 285 end; 286 287 writeln(f[ox,o[ox,0]].v); 288 289 close(input);close(output); 290 291 end.