Finding Palindromes （exkmp+trie）

https://cn.vjudge.net/problem/POJ-3376

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 li, then a single space and a string of li 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

3
1 a
2 ab
2 ba
      

Sample Output

5      

Hint

The 5 palindromes are: 

a a a ba ab a ab ba ba ab 

题意：

给定n个字符串。问把他们两两组合起来(包括自己和自己组合），能有多少个回文串。

字符串s与字符串t组合要想组成回文串

1.lens = lent,   s = rt  (rt为t的反串）

2.lens > lent,  rt 是 s 的前缀，且s剩下的部分是回文串

3.lens < lent,  s 是 rt 的前缀，且rt剩下的部分是回文串

可以先用一个扩展kmp，利用其extend数组求出每个串的最长回文后缀和最长回文前缀

把原串建一棵字典树，用反串匹配

book[0][i]表示原串剩下的部分是回文串

book[1][i]表示反串剩下的部分是回文串

num[i]表示以该节点为结尾的原串的数量

val[i]表示从i往后是回文串的数量

出题人太狠，时间和空间卡的太狼灭了

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn=2e6+5;
const int maxc=26;
char s[maxn],rs[maxn];
int l[maxn],r[maxn];
bool book[2][maxn];
int Next[maxn];//T[i]...T[m - 1]与T的最长相同前缀长度；
int extend[maxn];//S[i]...S[n - 1]与T的最长相同前缀长度。
/*求解T中Next[]，注释参考GetExtend()*/
void GetNext(char T[],int &m,int Next[]){
	int a=0,p=0;
	Next[0]=m;
	for(int i=1;i<m;i++){
		if(i>=p||i+Next[i-a]>=p){
			if(i>=p)
				p=i;
 
			while(p<m&&T[p]==T[p-i])
				p++;
 
			Next[i]=p-i;
			a=i;
		}else
			Next[i]=Next[i-a];
	}
}
 
/*求解extend[]*/
void GetExtend(char S[],char T[],int extend[],int Next[],int flag,int l){
	int a=0,p=0;
	int n=strlen(S);
    int m=strlen(T);
	GetNext(T,m,Next);
	for(int i=0;i<n;i++){
		if(i>=p||i+Next[i-a]>=p){//i>=p的作用：举个典型例子，S和T无一字符相同
			if(i>=p)
				p=i;
 
			while(p<n&&p-i<m&&S[p]==T[p-i])
				p++;
 
			extend[i]=p-i;
			a=i;
		}else
			extend[i]=Next[i-a];
	}
	for(int i=0;i<n;i++){
		if(i+extend[i]==n) book[flag][l+i]=true;
	}
}
int root=1;
int tot=0;
struct TRIE{
	int Next[maxn][maxc];
	int num[maxn];
	int val[maxn];
	int base;
	void init(){
		for(int i=0;i<=tot;i++) {
			memset(Next[i],0,sizeof(Next[i]));
			num[i]=0;
			val[i]=0;
		}
		tot=1;
	}
	void add(char s[],int now,int l){
		int len=strlen(s);
		for(int i=0;i<len;i++){
			int c=s[i]-base;
			val[now]+=book[0][l+i];
			if(!Next[now][c]){
			 Next[now][c]=++tot;
			} 
			now=Next[now][c];
		}
		num[now]++;
	}
	ll find(char s[],int len,int now,int l){
		ll ans=0;
		for(int i=0;i<len;i++){
			int c=s[i]-base;
			now=Next[now][c];
			if(!now) break;
			if((i<len-1&&book[1][l+i+1])||(i==len-1)) ans+=num[now];
		} 
		if(now) ans+=val[now];
		return ans; 
	}
}tr;
int main(){
    int n; 
    scanf("%d",&n);
    tr.init();
    tr.base='a';
    int sumlen=0;
	for(int i=1;i<=n;i++){
		int len;
		scanf("%d%s",&len,s+sumlen);
		int Len=sumlen+len;
		for(int j=0;j<len;j++){
			rs[sumlen+j]=s[sumlen+len-j-1];
		}
		l[i]=sumlen;
		r[i]=Len-1;
		sumlen=Len;
	    GetExtend(s+l[i],rs+l[i],extend,Next,0,l[i]);
	    GetExtend(rs+l[i],s+l[i],extend,Next,1,l[i]);
	    tr.add(s+l[i],1,l[i]);	
	}
	ll ans=0;
	for(int i=1;i<=n;i++) {
		ans+=tr.find(rs+l[i],r[i]-l[i]+1,1,l[i]);
	}
	printf("%lld\n",ans);
	return 0;
}