URAL 1960 Palindromes and Super Abilities (Palindromic Tree)

题目大意：

就是对于给出的字符串s (长度不超过10^5), 求出其各个前缀含有的不同回文串的数量

大致思路：

其实就是Palindromic Tree的裸题...看懂了Palindromic Tree的结构之后这个简直不能再明显, 当做联系Palindromic Tree来做了,

代码如下：

Result  :  Accepted     Memory  :  12222 KB     Time  :  187 ms

/*
* Author: Gatevin
* Created Time:  2015/3/30 20:24:10
* File Name: Rin_Tohsaka.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

#define maxn 100010
char s[maxn];

struct Palindromic_Tree
{
struct node
{
int next[26];
int len, sufflink;
lint cnt;
};
node tree[maxn];
int L, len, suff;
void newnode()
{
L++;
for(int i = 0; i < 26; i++)
tree[L].next[i] = -1;
tree[L].len = tree[L].sufflink = tree[L].cnt = 0;
return;
}
void init()
{
L = 0, suff = 2;
newnode(), newnode();
tree[1].len = -1, tree[1].sufflink = 1;
tree[2].len = 0, tree[2].sufflink = 1;
return;
}
bool addLetter(int pos)
{
int cur = suff, curlen = 0;
int alp = s[pos] - 'a';
while(1)
{
curlen = tree[cur].len;
if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
break;
cur = tree[cur].sufflink;
}
if(tree[cur].next[alp] != -1)
{
suff = tree[cur].next[alp];
return false;
}
newnode();
suff = L;
tree[L].len = tree[cur].len + 2;
tree[cur].next[alp] = L;
if(tree[L].len == 1)
{
tree[L].sufflink = 2;
tree[L].cnt = 1;
return true;
}
while(1)
{
cur = tree[cur].sufflink;
curlen = tree[cur].len;
if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
{
tree[L].sufflink = tree[cur].next[alp];
break;
}
}
tree[L].cnt = 1 + tree[tree[L].sufflink].cnt;
return true;
}
};

Palindromic_Tree pal;

int main()
{
scanf("%s", s);
pal.init();
int len = strlen(s);
lint ans = 0;
for(int i = 0; i < len; i++)
printf("%I64d ", ans += pal.addLetter(i));
return 0;
}