codeforces Gym 100548G - The Problem to Slow Down You 回文树

直接从两棵树的奇根和偶根dfs就可以了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
const int maxn=200100;
const int INF=1e9+10;

char s[maxn],t[maxn];
int ls,lt;

///PalinTree
struct PalinTree
{
    int ch[maxn][26],f[maxn];
    int tot,n,last;
    int s[maxn];
    int cnt[maxn],len[maxn];
    int L[maxn],R[maxn];
    int newnode(int l)
    {
        MS0(ch[tot]);
        cnt[tot]=0;
        len[tot]=l;
        return tot++;
    }
    void init()
    {
        tot=0;
        newnode(0);
        newnode(-1);
        n=last=0;
        s[n]=-1;f[0]=1;
        len[0]=0;len[1]=-1;
    }
    int get_fail(int x)
    {
        while(s[n-len[x]-1]!=s[n]) x=f[x];
        return x;
    }
    void add(int c)
    {
        c-='a';
        s[++n]=c;
        last=get_fail(last);
        if(!ch[last][c]){
            int cur=newnode(len[last]+2);
            f[cur]=ch[get_fail(f[last])][c];
            ch[last][c]=cur;
            R[cur]=n;
            L[cur]=n-len[cur]+1;
        }
        last=ch[last][c];
        cnt[last]++;
    }
    void calc()
    {
        for(int i=tot-1;i>=0;i--) cnt[f[i]]+=cnt[i];
    }
    void build(char *str)
    {
        init();
        int ls=strlen(str);
        REP(i,0,ls-1) add(str[i]);
    }
};PalinTree pt,ps;

ll dfs(int us,int ut)
{
    ll res=0;
    if(us!=0&&us!=1) res=1LL*ps.cnt[us]*pt.cnt[ut];
    REP(c,0,25){
        if(ps.ch[us][c]==0||pt.ch[ut][c]==0) continue;
        res+=dfs(ps.ch[us][c],pt.ch[ut][c]);
    }
    return res;
}

ll solve()
{
    ps.build(s+1);ps.calc();
    pt.build(t+1);pt.calc();
    return dfs(0,0)+dfs(1,1);
}

int main()
{
    freopen("in.txt","r",stdin);
    int T;cin>>T;int casen=1;
    while(T--){
        printf("Case #%d: ",casen++);
        scanf("%s%s",s+1,t+1);
        printf("%I64d\n",solve());
    }
    return 0;
}