POJ3267 The Cow Lexicon——动态规划——pku3267

简单的动态规划。状态转移方程：

f[i]=Min{f[i-1]+1,Make(i)}

其中Make为自定义函数：表示取了i这个位置上的字符之后前面至少要删掉多少个才能满足条件。

代码如下：

Program Lexicon;//By_Thispoet
Const
maxn=300;
Var
i,j,k,m,n				:Longint;
f						:Array[0..maxn]of Longint;
std						:Array[1..maxn*2]of String[25];
word					:Array[0..maxn]of Char;
temp					:Longint;

Function Min(i,j:Longint):Longint;
begin
if i<j then exit(i);exit(j);
end;

Function Make(i:Longint):Longint;
var j,k,p,q:Longint;
begin
make:=maxlongint;
for j:=1 to n do
begin
if std[j][length(std[j])]=word[i] then
begin
p:=0;q:=i;
k:=length(std[j]);
while(k>0)and(q>0)do
begin
while (std[j][k]<>word[q])and(q>0) do
begin
inc(p);dec(q);
end;
if std[j][k]=word[q] then
begin
dec(k);dec(q);
end;
end;
if k=0 then Make:=Min(Make,p+f[q]);
end;
end;
end;

BEGIN

readln(n,m);
for i:=1 to m do read(word[i]);readln;
for i:=1 to n do
readln(std[i]);

fillchar(f,sizeof(f),0);
for i:=1 to m do
begin
temp:=Make(i);
f[i]:=Min(f[i-1]+1,Temp);
end;

writeln(f[m]);

END.

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