HDU 2476 interval DP

Test instructions

Give two strings of S1 and S2, only one interval can be brushed once, ask at least a few times to let S1=S2

For example, zzzzzfzzzzz, with a length of 11, we'll take the subscript as 0~10

First brush the 0~10 once, become aaaaaaaaaaa

1~9 Brush once, Abbbbbbbbba

2~8:abcccccccba

3~7:abcdddddcba

4~6:abcdeeedcab

5:abcdefedcab

That's 6 times, it's a S2 string.

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <map>5#include <iostream>6#include <algorithm>7 using namespacestd;8 Const intmaxn= the;9 intDP[MAXN][MAXN];Ten CharA[MAXN],B[MAXN]; One intANS[MAXN]; A intMain () - { -Freopen ("1.in","R", stdin); the      while(cin>>a>>b) -     { -         inti,j,k,n,m,p,q,len,t; -n=strlen (a); +Memset (DP,0,sizeof(DP)); -         //for ease of handling, one needs to change in order to get the corresponding letter.  +         //because when a paragraph is updated, the values of the two strings that are originally equal are changed.  A          for(i=0; i<n;i++) dp[i][i]=1; at          for(len=2; len<=n;len++)//Enumeration Length -         { -              for(i=0; i<n-len+1; i++)//enumeration starting point -             { -j=i+len-1;//End -                 //cout<<i<< "" <<j<<endl; indp[i][j]=dp[i+1][j]+1;// !! Because only the comparison is b[i]=b[k], so can not be used dp[i][j]=dp[i][j-1]+1 -                  for(k=i+1; k<=j;k++)//Enumerate split points to                 { +                     if(b[i]==B[k]) -                     { the                         //cout<<i<< "" <<k<<endl; *Dp[i][j]=min (dp[i][j],dp[i+1][k]+dp[k+1][j]); $                     }Panax Notoginseng                 } -              } the          } +          for(i=0; i<n;i++) Aans[i]=dp[0][i]; the          for(i=0; i<n;i++) +         { -            //recalculates the same point in two strings as the characters.  $            if(a[i]==B[i]) $             { -                 if(i==0) ans[i]=0; -                 Elseans[i]=ans[i-1]; the             } -             ElseWuyi             { the                  for(j=0; j<i;j++) -Ans[i]=min (ans[i],ans[j]+dp[j+1][i]); Wu             } -         } Aboutcout<<ans[n-1]<<Endl; $     } -     return 0; -}

HDU 2476 interval DP