【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=1000000007;
var dp:array[0..50,0..500,0..500]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[0,0,0]:=1;
 for i:=1 to n do
  if a[i]=b[1] then dp[1,i,1]:=1;
 for i:=1 to k1 do
  for j:=1 to n do
   for k:=1 to m do
   begin
    if a[j]<>b[k] then continue;
    if a[j-1]<>b[k-1] then
     for x:=1 to j-1 do dp[i,j,k]:=(dp[i,j,k]+dp[i-1,x,k-1]) mod mo
      else
      begin
       for x:=1 to j-1 do dp[i,j,k]:=(dp[i,j,k]+dp[i-1,x,k-1]) mod mo;
       dp[i,j,k]:=(dp[i,j,k]+dp[i,j-1,k-1]) mod mo;
      end;
   end;
 for i:=1 to n do ans:=(ans+dp[k1,i,m]) mod mo;
 writeln(ans);
// close(input);
// close(output);
end.

 

加了优化

const mo=1000000007;
var dp,f:array[0..1,0..1000,0..200]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[0,0,0]:=1;
 for i:=0 to n do f[0,i,0]:=1;
 for i:=1 to k1 do
 begin
  v:=1-v;
  fillchar(f[v],sizeof(f[v]),0);
  fillchar(dp[v],sizeof(dp[v]),0);
  for j:=1 to n do
   for k:=1 to m do
   begin
    if a[j]=b[k] then
    begin
     dp[v,j,k]:=f[1-v,j-1,k-1];
     if a[j-1]=b[k-1] then dp[v,j,k]:=(dp[v,j,k]+dp[v,j-1,k-1]) mod mo;
    end;
    f[v,j,k]:=(f[v,j-1,k]+dp[v,j,k]) mod mo;
   end;
 end;
 for i:=1 to n do ans:=(ans+dp[v,i,m]) mod mo;
 writeln(ans);
 //close(input);
 //close(output);
end.