UVALive - 7041 G - The Problem to Slow Down You

题意：求两个串的公共回文子串个数

题解:建两个回文自动机，从0和1各跑一边就是答案了，因为对于回文自动机来说，从头开始dfs就能找出该字符串的所有回文串

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=400000+10,maxn=400000+10,inf=0x3f3f3f3f;

struct PAM{
    int ch[N][26],fail[N],cnt[N],len[N],s[N];
    int last,n,p;
    int newnode(int w)
    {
        for(int i=0;i<26;i++)ch[p][i] = 0;
        cnt[p] = 0;
        len[p] = w;
        return p++;
    }
    void init()
    {
        p = last = n = 0;
        newnode(0);
        newnode(-1);
        s[n] = -1;
        fail[0] = 1;
    }
    int getfail(int x)
    {
        while(s[n-len[x]-1] != s[n]) x = fail[x];
        return x;
    }
    void add(int c)
    {
        s[++n] = c;
        int cur = getfail(last);
        if(!ch[cur][c]){
            int now = newnode(len[cur]+2);
            fail[now] = ch[getfail(fail[cur])][c];
            ch[cur][c] = now;
        }
        last = ch[cur][c];
        cnt[last]++;
    }
    void cal()
    {
        for(int i=p-1;i>=0;i--)
        {
            cnt[fail[i]]+=cnt[i];
//            printf("%d %d\n",i,cnt[i]);
        }
//        puts("------------");
    }
}pam1,pam2;
ll ans;
void dfs(int u1,int u2)
{
    if(u1!=0&&u1!=1)ans+=1ll*pam1.cnt[u1]*pam2.cnt[u2];
//    printf("%d %d\n",pam1.cnt[u1],pam2.cnt[u2]);
    for(int i=0;i<26;i++)
        if(pam1.ch[u1][i]&&pam2.ch[u2][i])
            dfs(pam1.ch[u1][i],pam2.ch[u2][i]);
}
char s[N];
int main()
{
    int T;scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        pam1.init();pam2.init();
        scanf("%s",s);int n=strlen(s);
        for(int i=0;i<n;i++)pam1.add(s[i]-'a');
        scanf("%s",s);n=strlen(s);
        for(int i=0;i<n;i++)pam2.add(s[i]-'a');
        pam1.cal();pam2.cal();
        ans=0;dfs(0,0);dfs(1,1);
        printf("Case #%d: %lld\n",cas,ans);
    }
    return 0;
}
/********************
3
abacab
abccab
faultydogeuniversity
hasnopalindromeatall
abbacabbaccab
youmayexpectedstrongsamplesbutnow
********************/