【NOIP2015】子串（字符串DP）

题意：有AB两个字符串，用A中连续的K串匹配B全串，问不同的方案总数

n<=1000,m<=200,k<=m

思路：设dp[k,i,j]为用k串 A中前i个字符匹配B中前j个字符的方案总数

首先dp[k,i,j]=0 (a[i]<>b[j])

然后就是考虑dp[k,i,j]能否从dp[k,i-1,j-1]即前一个连续转移来 dp[k,i,j]=dp[k,i,j]+dp[k,i-1,j-1]

还有就是另起一串 dp[k,i,j]=dp[k-1,1,j-1]+dp[k-1,2,j-1]+...+dp[k-1,i-1,j-1]

转移用前缀和优化，空间用滚动数组优化

暴力

 const mo=;
 var dp:array[..,..,..]of longint;
     a,b:ansistring;
     n,m,k1,i,j,k,x,ans:longint;

 begin
  //assign(input,'1.in'); reset(input);
  //assign(output,'1.out'); rewrite(output);
  readln(n,m,k1);
  readln(a);
  readln(b);
  dp[,,]:=;
  for i:= to n do
   if a[i]=b[] then dp[,i,]:=;
  for i:= to k1 do
   for j:= to n do
    for k:= to m do
    begin
     if a[j]<>b[k] then continue;
     if a[j-]<>b[k-] then
      for x:= to j- do dp[i,j,k]:=(dp[i,j,k]+dp[i-,x,k-]) mod mo
       else
       begin
        for x:= to j- do dp[i,j,k]:=(dp[i,j,k]+dp[i-,x,k-]) mod mo;
        dp[i,j,k]:=(dp[i,j,k]+dp[i,j-,k-]) mod mo;
       end;
    end;
  for i:= to n do ans:=(ans+dp[k1,i,m]) mod mo;
  writeln(ans);
 // close(input);
 // close(output);
 end.

加了优化

 const mo=;
 var dp,f:array[..,..,..]of longint;
     a,b:ansistring;
     n,m,k1,i,v,j,k,ans:longint;

 begin
  //assign(input,'1.in'); reset(input);
  //assign(output,'1.out'); rewrite(output);
  readln(n,m,k1);
  readln(a);
  readln(b);
  dp[,,]:=;
  for i:= to n do f[,i,]:=;
  for i:= to k1 do
  begin
   v:=-v;
   fillchar(f[v],sizeof(f[v]),);
   fillchar(dp[v],sizeof(dp[v]),);
   for j:= to n do
    for k:= to m do
    begin
     if a[j]=b[k] then
     begin
      dp[v,j,k]:=f[-v,j-,k-];
      if a[j-]=b[k-] then dp[v,j,k]:=(dp[v,j,k]+dp[v,j-,k-]) mod mo;
     end;
     f[v,j,k]:=(f[v,j-,k]+dp[v,j,k]) mod mo;
    end;
  end;
  for i:= to n do ans:=(ans+dp[v,i,m]) mod mo;
  writeln(ans);
  //close(input);
  //close(output);
 end.