美團點評A輪B題

合併迴文子串

題目鏈接
題意是給兩個字符串,問合併之後可以得到的最長迴文串有多長.
如dba abcd 合併後可以成爲 dabcbad 最長迴文串即爲它本身.
求一串字符串的最長迴文串的dp解法是dp[i][j]爲1表示i到j爲合法迴文串,爲0表示不合法,當(s[i] == s[j]&& dp[i+1][j-1]) dp[i][j] = 1, 空串的時候也是合法的.
兩個字符串的做法在一串的基礎上加兩維狀態即可,即dp[i][j][k][l]表示A串的i到j且B串的k到l是否爲合法迴文串.注意把空串的情況處理一下

#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<time.h>
#include<cstdio>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<iostream>
#include<stdlib.h>
using namespace std;
#define  LONG long long
const LONG  INF=0x3f3f3f3f;
const LONG  MOD=1e9+ 7;
const double PI=acos(-1.0);
#define clrI(x) memset(x,-1,sizeof(x))
#define clr0(x) memset(x,0,sizeof x)
#define clr1(x) memset(x,INF,sizeof x)
#define clr2(x) memset(x,-INF,sizeof x)
#define EPS 1e-10
#define lson  l , mid , rt<< 1
#define rson  mid + 1 ,r , (rt<<1)+1
#define root 1, m , 1
int dp[53][53][53][53] ;
char A[60] , B[60] ;
int main()
{
    int T;
    cin >>T;
    while( T -- )
    {
        clr0(dp) ;
        A[0] = '#' , B[0] = '#' ;
        scanf("%s%s", A+1 , B +1 ) ;
        int Alen = strlen(A) - 1;
        int Blen = strlen(B) - 1;
        for( int i = 1 ; i <= Alen+1 ; ++i)
        for(int j = 1 ;j <= Blen +1; ++ j)
        dp[i][i-1][j][j-1] = 1;
        for(int k = 1; k <= Blen +1; ++ k)
        for(int len1 = 0; len1 <= Alen ; ++len1)
        {
            for(int i = 1 ; i <= Alen ; ++ i)
            {
                int j = i +len1 ;
                if( j > Alen )
                break  ;
                if(len1 == 0)
                {

                dp[i][j][k][k-1] =  1;
                continue ;
                }

                if(A[i] == A[j] && dp[i+1][j-1][k][k-1])
                    dp[i][j][k][k-1] = 1;
            }
        }
        for(int k  =1 ; k <= Alen + 1; ++ k)
        for(int len1 = 0; len1 <= Blen ; ++len1)
        {
            for(int i = 1 ; i <= Blen ; ++ i)
            {
                int j = i +len1 ;
                if( j > Blen )
                break  ;
                if(len1 == 0)
                {
                    dp[k][k-1][ i ][j] =  1;
                    continue ;
                }

                if(B[i] == B[j] && dp[k][k-1][i+1][j-1])
                dp[k][k-1][i][j] = 1;

            }
        }
        int ans = 0 ;
        for(int len1 = 0; len1 <= Alen ; ++ len1)
        {
            for(int i = 1; i <= Alen ; ++ i)
            {
                int j = i + len1 ;
                if( j > Alen) break ;
                for(int len2 = 0 ; len2 <= Blen ; ++len2 )
                {
                    for(int k = 1; k<= Blen ; ++k)
                    {
                        int l = k + len2 ;
                        if(l > Blen) break ;
                        if(B[k] == A[j] && dp[i][j-1][k+1][l])
                        dp[i][j][k][l] = 1 ;
                        if(B[l] == A[i] && dp[i+1][j][k][l-1])
                        dp[i][j][k][l] = 1 ;
                        if(A[i] == A[j] && i != j && dp[i+1][j-1][k][l])
                        dp[i][j][k][l] = 1;
                        if(B[k] == B[l] && k != l && dp[i][j][k+1][l-1])
                        dp[i][j][k][l] = 1 ;
                        if(dp[i][j][k][l])
                        ans = max(ans , j - i + 1 + l - k + 1) ;
                    }
                }
            }
        }
        printf("%d\n",ans) ;
    }
    return 0 ;
}