題意：給你兩個字符串，問你有多少對公共回文串。

　　思路：先對a字符串建回文樹。然後再把b字符串加進去就好了。

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
#include<queue>
#include<set>
#include<string>
#include<map>
#include 
 
<time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 400005+1000;
const int sigma=26;
const ll mod = 1000000007;
const int INF = 0x3f3f3f;
const db eps = 1e-9;
struct ptree{
    char s[maxn];
    int next[maxn][sigma], fail[maxn], len[maxn];
    ll cntt[maxn], cnt[maxn];
    
 
int last, n, p;
    ll res;
    inline int newnode(int l) {
        memset(next[p], 0, sizeof(next[p]));
        cnt[p]=0;
        cntt[p]=0;
        len[p]=l;
        return p++;
    }
    inline void init() {
        n=0, p=0, last=0;
        newnode(0), newnode(-1);
        s[n]=-1;
        fail[0]=1;
    }
    
 
void initt() {
        n=0, last=0;
        s[0]=-1;
        fail[0]=1;
    }
    inline int FL(int x) {
        while(s[n-len[x]-1]!=s[n])  x=fail[x];
        return x;
    }
    void add(char c) {
        c-=‘a‘;
        s[++n]=c;
        int cur=FL(last);
        if (!next[cur][c]) {
            int now=newnode(len[cur]+2);
            fail[now]=next[FL(fail[cur])][c];
            next[cur][c]=now;
        }
        last=next[cur][c];
        ++cnt[last];
    }
    inline ll countt() {
        for (int i=p-1; ~i; --i) {
            cnt[fail[i]]+=cnt[i];
        }
    }
    void add1(char c) {
        c-=‘a‘;
        s[++n]=c;
        int cur=FL(last);
        if (!next[cur][c]) {
            int now=newnode(len[cur]+2);
            fail[now]=next[FL(fail[cur])][c];
            next[cur][c]=now;
        }
        last=next[cur][c];
        ++cntt[last];
    }
    inline ll countt1() {
        for (int i=p-1; ~i; --i) {
            cntt[fail[i]]+=cntt[i];
        }
    }
    ll cal() {
        //cnt是a字符串的個數，cntt是a相同的回文串的個數
        ll pk=0;
        for (int i=2; i<p; i++) {
            pk += cnt[i]*cntt[i];
        }
        return pk;
    }
}p;
char a[maxn], b[maxn];
void solve(){
    scanf("%s", a);
    int len=strlen(a);
    scanf("%s", b);
    int lenn=strlen(b);
    p.init();   //對a建回文樹
    for (int i=0; i<len; i++) {
        p.add(a[i]);
    }
    p.countt();
    p.initt();  //再次初始化
    for (int i=0; i<lenn; i++) {
        p.add1(b[i]);
    }
    p.countt1();
    ll ans=p.cal();
    static int pa=1;
    printf("Case #%d: %lld\n", pa++, ans);
}
int main() {
    int t = 1;
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    scanf("%d", &t);
    while(t--) {
        solve();
    }
    return 0;
}

UVALive-7041(回文樹