2014-2015 ACM-ICPC, Asia Xian Regional Contest(G - The Problem to Slow Down You-回文树)

求两个字符串的公共回文子序列个数

建立2个回文树，统计2个字符串每个本质不同的回文串个数，
然后同时遍历2个回文树（取交），或者hash什么的都行

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p]) 
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F1 (3000000009uLL)
#define F2 (1000000000000000000000000000000uLL) 
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define MAXN (300000+10)
typedef long long ll;
typedef unsigned long long ull;
const ull base=300007uLL;
const ull base2=500009uLL;

map<pair<ull,ull> ,int> h;
map<pair<ull,ull> ,int>::iterator it;

namespace Palindromic_Tree {
    int s[MAXN],n;
    int tot,next[MAXN][26],link[MAXN],len[MAXN],last;
    int cnt[MAXN];

    int newnode(int l)
    {
        len[tot]=l;
        return tot++;
    }
    void mem(int N) {
        Rep(i,N) cnt[i]=0,MEM(next[i]);
// MEM(s) MEM(next) MEM(link) MEM(len) MEM(cnt)
        n=tot=0;
        newnode(0); newnode(-1);
        link[0]=link[1]=1; s[0]=27;
        last=0;
    }

    int getnode(int x)
    {
        while (s[ n - len[x]-1 ]  != s[n] ) x=link[x];
        return x; 
    }

    void add(int c) {
        s[++n]=c;
        int cur=getnode(last);
        if (!next[cur][c])
        {
            int now=newnode(len[cur]+2);
            int tmp=getnode(link[cur]);
            link[now]=next[tmp][c];
            next[cur][c] = now;

        }       
        last=next[cur][c];
        cnt[last]++;
    }

    ull h1[MAXN],h2[MAXN];
    void dfs()
    {
        Rep(x,tot) {
            ull p1=h1[x],p2=h2[x];
            Rep(c,26) {
                if (next[x][c]) h1[next[x][c]]=(p1*base+1+c)%F1,h2[next[x][c]]=(p2*base2+1+c)%F2; 
            } 
            if (x==0||x==1) ; 
            else{
                h[mp(h1[x],h2[x])]=cnt[x];  

            }
         } 
    }

    void work()
    {
        RepD(i,tot-1) cnt[link[i]]+=cnt[i];
        h1[0]=h2[0]=21,h1[1]=h2[1]=12;
        dfs();

    }
    bool b[MAXN];
    ll _ans;
    void dfs2()
    {
        Rep(x,tot) b[x]=0; b[0]=b[1]=1; 
        Rep(x,tot) {
            if (!b[x]) continue;
            ull p1=h1[x],p2=h2[x];
            it = h.find(mp(p1,p2));
            if (x>1&&it==h.end()) continue ;
                else if (x>1) {
                    _ans+=(ll)(*it).se*(ll)cnt[x]; 
// cout<< (*it).se <<' '<< cnt[x] <<endl;
                }
                Rep(c,26) {
                    if (next[x][c]) 
                    {

                        b[next[x][c]]=1;

                        h1[next[x][c]]=(p1*base+1+c)%F1,h2[next[x][c]]=(p2*base2+1+c)%F2; 
                    }
            }  

        } 
    }

    void work2()
    { 
        RepD(i,tot-1) cnt[link[i]]+=cnt[i];
        h1[0]=h2[0]=21,h1[1]=h2[1]=12;
        _ans=0;
        dfs2();
        cout<<_ans<<endl;
    }


}
using namespace Palindromic_Tree;

char S[MAXN];
int N;
int main()
{
// freopen("CF100548G_data.in","r",stdin);
// freopen(".out","w",stdout);

    int T; cin>>T;
    For(kcase,T)
    { 
        h.clear();
        scanf("%s",S);
        int N=strlen(S);
        Palindromic_Tree::mem(N+6);
        Rep(i,N) Palindromic_Tree::add(S[i]-'a');

        Palindromic_Tree::work(); 

        scanf("%s",S);
        N=strlen(S);
        Palindromic_Tree::mem(N+6);
        Rep(i,N) Palindromic_Tree::add(S[i]-'a');

        printf("Case #%d: ",kcase);
        Palindromic_Tree::work2(); 


    }
    return 0;
}