128 条题解
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Plam_gxy LV 10 @ 2009-07-18 11:04:35
program b_max;
type
gz=array[1..6,1..2]of string;
ww=array[1..10000]of record
now:string;
step:longint;
end;
kkk=array[0..10000]of longint;
var
a:gz;
dui1,dui2:ww;
qishi,mubiao:string;
i,j,n,k,lena:longint;
data,h1,t1,m1,t2,h2,m2:longint;
posi1,posi2:kkk;procedure fill;
var s1,s2,s3:string;
k:longint;
begin
readln(s1);
k:=pos(' ',s1);
qishi:=copy(s1,1,k-1);
mubiao:=copy(s1,k+1,length(s1)-k);
while not eof do
begin
readln(s1);
k:=pos(' ',s1);
inc(lena);
a[lena,1]:=copy(s1,1,k-1);
a[lena,2]:=copy(s1,k+1,length(s1)-k);
end;
h1:=1; t1:=1; m1:=1;
h2:=1; t2:=1; m2:=1;
dui1[t1].now:=qishi;
dui2[t2].now:=mubiao;
end;procedure search(var posi1:kkk;s1,s2:string);
var i,j,l1,l2:longint;
begin
fillchar(posi1,sizeof(posi1),0);
l1:=length(s1);
l2:=length(s2);
for i:=1 to l1 do
if copy(s1,i,l2)=s2 then
begin
inc(posi1[0]);
posi1[posi1[0]]:=i;
end;
end;procedure change1(x,y:longint);
var i,j:longint;
s1,s2:string;
begin
s1:=copy(dui1.now,1,y-1);
s2:=copy(dui1.now,y+length(a[x,1]),length(dui1.now)-y-length(a[x,1])+1);
inc(t1);
dui1[t1].now:=s1+a[x,2]+s2;
dui1[t1].step:=dui1.step+1;
end;procedure change2(x,y:longint);
var i,j:longint;
s1,s2:string;
begin
s1:=copy(dui2.now,1,y-1);
s2:=copy(dui2.now,y+length(a[x,2]),length(dui2.now)-y-length(a[x,2])+1);
inc(t2);
dui2[t2].now:=s1+a[x,1]+s2;
dui2[t2].step:=dui2.step+1;
end;function check1:boolean;
var i:longint;
begin
for i:=t2 downto 1 do
if dui2[i].now=dui1[t1].now then
begin
data:=dui2[i].step;
exit(true);
end;
exit(False);
end;function check2:boolean;
var i:longint;
begin
for i:=t1 downto 1 do
if dui1[i].now=dui2[t2].now then
begin
data:=dui1[i].step;
exit(true);
end;
exit(false);
end;begin
fill;
repeat
while h1m1+1 do
begin
for i:=1 to lena do
begin
search(posi1,dui1.now,a -
0
var
a,b:string;
n,answer:integer;
len:array[0..5]of integer;
s1,s2:array[0..5]of string;
procedure reads(var s1,s2:string);
var
s:string;
begin
readln(s);
s1:=copy(s,1,pos(' ',s)-1);
s2:=copy(s,pos(' ',s)+1,length(s)-pos(' ',s));
end;
function max(a,b:integer):integer;
begin
if a>b then max:=a
else max:=b;
end;
procedure search(depth:integer;s:string;last:integer);
var
i,j,l:integer;
begin
if (depth=0) then
begin
if (s=b) then
begin
writeln(answer);
halt;
end;
exit;
end;
l:=length(s);
dec(depth);
for i:=1 to n do
for j:=max(1,last-len[i]+1) to l-len[i]+1 do
if (copy(s,j,len[i])=s1[i]) then
search(depth,copy(s,1,j-1)+s2[i]+copy(s,j+len[i],l-j-len[i]+1),j);
end;
begin
reads(a,b);
while (not seekeof) do
begin
inc(n);
reads(s1[n],s2[n]);
len[n]:=length(s1[n]);
end;
for answer:=1 to 10 do
search(answer,a,1);
writeln('NO ANSWER!');
end. -
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mingyueyucai LV 3 @ 2008-11-11 08:31:36
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms双搜，不用hash.
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CSAPC_1_1636 LV 3 @ 2008-11-07 21:59:27
迭代加深+哈西判重就可以过了，而且程序写的很舒服。
我一直觉得双向广搜写的很难受…… -
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WiDe_Courage LV 4 @ 2008-11-05 15:16:14
不会2向广搜 但是貌似dfs就行
跟luyifan的代码差不多 -
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norsonerror LV 4 @ 2008-10-27 20:01:39
不知道什么叫hash
其实刚知道什么叫广搜
不过还是过了
第120道题,一遍AC
庆祝一下
Plam_gxy
LV 10
@ 2009-07-18 11:04:35
