不难想到是一个布尔型dp，
不难想到用f[i,j,k]表示区间[i,j]能否变为字母k
不难想到对于f[i,j,k]，拆[i,j]成两个区间，然后穷举k的每一个变换来判断
感觉记忆化搜索写的比较顺，就写了记忆化
1 const a:array[1..4] of char=('W','I','N','G'); 2 3 var b:array[0..300] of longint; 4 f:array[0..300,0..300,0..4] of integer; 5 w:array[0..4,0..20] of longint; 6 c:array[1..5] of longint; 7 l,i,j,k,e,n,m,p:longint; 8 flag:boolean; 9 s:string; 10 11 function exc(x:char):longint; 12 begin 13 if x='W' then exit(1) 14 else if x='I' then exit(2) 15 else if x='N' then exit(3) 16 else if x='G' then exit(4); 17 end; 18 19 function search(l,r,k:longint):integer; 20 var i,j,x1,x2:longint; 21 begin 22 if f[l,r,k]<>0 then exit(f[l,r,k]); 23 if l=r then 24 begin 25 f[l,l,b[l]]:=1; 26 if k=b[l] then exit(1) 27 else begin 28 f[l,l,k]:=-1; 29 exit(-1); 30 end; 31 end; 32 for i:=1 to c[k] do 33 begin 34 x1:=(w[k,i]-1) div 4+1; 35 x2:=(w[k,i]-1) mod 4+1; 36 for j:=l to r-1 do 37 begin 38 if (search(l,j,x1)=1) and (search(j+1,r,x2)=1) then 39 begin 40 f[l,r,k]:=1; 41 exit(1); 42 end; 43 end; 44 end; 45 f[l,r,k]:=-1; 46 exit(-1); 47 end; 48 49 begin 50 for i:=1 to 4 do 51 read(c[i]); 52 readln; 53 for i:=1 to 4 do 54 for j:=1 to c[i] do 55 begin 56 readln(s); 57 p:=(exc(s[1])-1)*4+exc(s[2]); 58 w[i,j]:=p; 59 end; 60 61 readln(s); 62 n:=length(s); 63 64 for i:=1 to n do 65 b[i]:=exc(s[i]); 66 67 flag:=false; 68 for i:=1 to 4 do 69 if search(1,n,i)=1 then 70 begin 71 write(a[i]); 72 flag:=true; 73 end; 74 if not flag then writeln('The name is wrong!') else writeln; 75 end.