回文树(统计所有回文串的个数) - MCCME 1750 Подпалиндромы

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

[b]Mean:[/b]

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)。

[b]analyse:[/b]

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。

[b]Time complexity: O(N)[/b]

[b]Source code: [/b]

/*
* 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;
}