回文自動機+DFS - The Problem to Slow Down You

題目來源

[The 2014 ACM-ICPC Asia Xi’an Regional Contest Problem G]

題目描述

給你兩個字元串，求這兩個字元串相同回文串的比對對數。

思路分析

每個字元串建一棵回文樹，分别從0結點和1結點兩棵樹一起往下dfs，對于同一條路徑上的結點，一定是相同的回文，然後兩個的數量相乘加到answer中。

時間複雜度

O(N)

代碼

/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-22-13.43
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define  LL long long
#define  ULL unsigned long long
using namespace std;

const int MAXN = 200050 ;
const int N = 26;
LL ans;
char s[MAXN];
struct Palindromic_Tree
{
     int next[MAXN][N];
     int fail[MAXN];
     int cnt[MAXN];
     int num[MAXN];
     int len[MAXN];
     int S[MAXN];
     int last;
     int n ;
     int p ;
     int newnode(int l)
     {
           for(int i = 0 ; i < N ; ++ i) next[p][i] = 0 ;
           cnt[p] = 0 ;
           num[p] = 0 ;
           len[p] = l ;
           return p ++ ;
     }
     void init()
     {
           p = 0 ;
           newnode(0) ;
           newnode(-1) ;
           last = 0 ;
           n = 0 ;
           S[n] = -1 ;
           fail[0] = 1 ;
     }
     int get_fail(int x)
     {
           while(S[n - len[x] - 1] != S[n]) x = fail[x] ;
           return x ;
     }
     void add(int c)
     {
           c -= 'a';
           S[++ n] = c ;
           int cur = get_fail(last);
           if(!next[cur][c])
           {
                 int now = newnode(len[cur] + 2);
                 fail[now] = next[get_fail(fail[cur])][c] ;
                 next[cur][c] = now;
                 num[now] = num[fail[now]] + 1 ;
           }
           last = next[cur][c];
           cnt[last] ++;
     }
     void count() { for(int i = p - 1 ; i >= 0 ; -- i) cnt[fail[i]] += cnt[i]; }
} t1,t2;

void dfs(int u,int v)
{
     for(int i=0; i<N; ++i)
     {
           int x=t1.next[u][i];
           int y=t2.next[v][i];
           if(x&&y)
           {
                 ans+=(LL)t1.cnt[x]*t2.cnt[y];
                 dfs(x,y);
           }
     }
}

int main()
{
     int t;
     scanf("%d",&t);
     for(int Cas=1; Cas<=t; ++Cas)
     {
           t1.init(),t2.init();
           scanf("%s",s);
           for(int i=0; s[i]; ++i) t1.add(s[i]);
           scanf("%s",s);
           for(int i=0; s[i]; ++i) t2.add(s[i]);
           t1.count(),t2.count();
           ans=0;
           dfs(0,0),dfs(1,1);
           printf("Case #%d: %lld\n",Cas,ans);
     }
     return 0;
}      

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