Finding Palindromes

Time Limit: 10000MS		Memory Limit: 262144K
Case Time Limit: 2000MS

Description

A word is called a palindrome if we read from right to left is as same as we read from left to right. For example, "dad", "eye" and "racecar" are all palindromes, but "odd", "see" and "orange" are not palindromes.

Given n strings, you can generate n × n pairs of them and concatenate the pairs into single words. The task is to count how many of the so generated words are palindromes.

Input

The first line of input file contains the number of strings n. The following n lines describe each string:

The i+1-th line contains the length of the i-th string l_i, then a single space and a string of l_i small letters of English alphabet.

You can assume that the total length of all strings will not exceed 2,000,000. Two strings in different line may be the same.

Output

Print out only one integer, the number of palindromes.

Sample Input

31 a2 ab2 ba

Sample Output

5

Hint

The 5 palindromes are: 

a

a a

ba ab

a ab

ba ba

ab 

 

建立trie树

用原串的反串在trie树上匹配

假设反串匹配到i，trie树上到j

1、如果j是单词节点，即以j结尾的单词是反串的前缀，那么如果i以后（反串剩余的部分，不包括i）是回文串，以j结尾的单词都可以与匹配的单词构成回文串

2、如果反串匹配完了，即反串是以i结尾的单词的前缀，那么j后面（不包括j）有多少回文串，就可以产生多少合法答案

关键：字符串的前缀回文和后缀回文

用扩展kmp

#include
      
       #include
       
        #include
        
         #define N 2000008using namespace std;bool f[2][N];long long ans;char T[N],S[N];int len,tot,root,id;int st[N],ed[N],cnt;int nxt[N],expand[N];int trie[N][26],mark[N],sum[N];void getnxt(char *s,int ll,int rr){    int a=ll;    nxt[0]=rr-ll+1;    while(a+1<=rr && s[a]==s[a+1]) a++;    nxt[1]=a-ll;    a=1+ll;    int p,l,j;    for(int k=2+ll;k<=rr;k++)    {        p=a-ll+nxt[a-ll]-1; l=nxt[k-a];        if(k-ll+l-1>=p)        {            j=p-k+ll+1>0 ? p-k+ll+1 : 0;            while(k+j<=rr && s[k+j]==s[j+ll]) j++;            nxt[k-ll]=j;            a=k;        }        else nxt[k-ll]=l;    }}void exkmp(char *s,char *t,int ll,int rr,int w){    getnxt(t,ll,rr);    int a=ll;    while(a<=rr && s[a]==t[a]) a++;    expand[0]=a-ll;    a=ll;    int p,l,j;    for(int k=ll+1;k<=rr;k++)    {        p=a-ll+expand[a-ll]-1; l=nxt[k-a];        if(k-ll+l-1>=p)        {            j=p-k+ll+1>0 ? p-k+ll+1 : 0;            while(k+j<=rr && s[k+j]==t[j+ll]) j++;            expand[k-ll]=j;            a=k;        }        else expand[k-ll]=l;    }    for(int i=ll-ll;i<=rr-ll;i++)     if(i+expand[i]==rr-ll+1) f[w][i+ll]=true;}void insert(int ll,int rr){    root=0;    for(int i=ll;i<=rr;i++)    {        id=S[i]-'a';        sum[root]+=f[0][i];        if(!trie[root][id]) trie[root][id]=++tot;        root=trie[root][id];    }    mark[root]++;}void find(int ll,int rr){    root=0;    for(int i=ll;i<=rr;i++)    {        id=T[i]-'a';        root=trie[root][id];        if(!root) return;        if(i!=rr&&f[1][i+1] || i==rr) ans+=mark[root];    }     ans+=sum[root];}int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        scanf("%d%s",&len,S+cnt);        for(int j=0;j