uva11404

删去字符串S中的一些字符，使剩下的字符组成最长的回文串（顺序连接）

 

区间dp f[i][j] 

f[i][j] = f[i+1][j-1]    i==j

 　　=min( f[i+1][j] ,f[i][j-1] )

 

#include<iostream>
#include<string>
#include <cstring>
#include <algorithm>
using namespace std;
 const int N=1005;
 int f[N][N],n;
 string g[N][N];
 char s[N];
 
 void solve(){
     int i,j;
     n=strlen(s+1);
     
     for(i=0;i<=n;i++)
      for(j=0;j<=n;j++){
           f[i][j]=0; g[i][j]="";
      }
     for(i=1;i<=n;i++) f[i][i]=0,g[i][i]=s[i];
     
     for(i=n;i>0;i--)
      for(j=i;j<=n;j++){
              if(i==j) continue;
              if(s[i]==s[j]){
                  f[i][j]=f[i+1][j-1];
                  g[i][j]=s[i]+g[i+1][j-1]+s[i];
              }
             else{
                 if(f[i][j-1]<f[i+1][j]){
                     f[i][j]=f[i][j-1]+1; g[i][j]=g[i][j-1];
                 }
                 else if(f[i][j-1]>f[i+1][j]){
                     f[i][j]=f[i+1][j]+1; g[i][j]=g[i+1][j];
                 }
                 else{
                     f[i][j]=f[i+1][j]+1;
                     g[i][j]=min(g[i+1][j],g[i][j-1]);
                 }
             }
        }
     cout<<g[1][n]<<endl;
 }
 int main(){
     while(cin>>s+1)
          solve();
 }