回文树(统计所有回文串的个数)

Подпалиндромы 

Problem's Link:  http://informatics.mccme.ru//mod/statements/view.php?chapterid=1750#

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Mean: 

给你一个长度不超过1e5的字符串，要统计总共有多少个回文串。(第一次刷俄语题，还好有google翻译)。

analyse:

如果题目说是不同回文串的话，这题就是一个裸的回文树。

由于要求所有子串的回文数，所以我们在插入的时候还需要使用suffixLink往上走统计答案。

Time complexity: O(N)

 

Source code: 

/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-17-19.13
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define  LL long long
#define  ULL unsigned long long
using namespace std;

const int MAXN = 105000;
struct node
{
     int next[26];
     int len;
     int sufflink;
     int num;
};
int len;
char s[MAXN];
node tree[MAXN];
int num; // node 1 - root with len -1, node 2 - root with len 0
int suff; // max suffix palindrome
long long ans;
bool addLetter(int pos)
{
     int cur = suff, curlen = 0;
     int let = s[pos] - 'a';
     while(true)
     {
           curlen = tree[cur].len;
           if(pos-1-curlen>=0&&s[pos-1-curlen]==s[pos]) break;
           cur = tree[cur].sufflink;
     }
     if(tree[cur].next[let])
     {
           suff = tree[cur].next[let];
           return false;
     } suff = ++num;
     tree[num].len = tree[cur].len + 2;
     tree[cur].next[let] = num;
     if(tree[num].len == 1)
     {
           tree[num].sufflink = 2;
           tree[num].num = 1;
           return true;
     }
     while(true)
     {
           cur = tree[cur].sufflink;
           curlen = tree[cur].len;
           if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
           {
                 tree[num].sufflink = tree[cur].next[let];
                 break;
           }
     }
     tree[num].num = 1 + tree[tree[num].sufflink].num;
     return true;
}
void initTree()
{
     num = 2; suff = 2;
     tree[1].len = -1; tree[1].sufflink = 1;
     tree[2].len = 0; tree[2].sufflink = 1;
}
int main()
{
     gets(s);
     initTree();
     for(int i = 0;s[i]; i++)
     {
           addLetter(i);
           ans += tree[suff].num;
     } cout << ans << endl;
     return 0;
}