URAL 2040 Palindromes and Super Abilities 2 (Palindromic Tree)

题目大意：

对于输入的字符串s (长度不超过5000000, 只包含a, b两种字符) 如果这个字符串从左到右依次加入形成的s, 当每个字符加入时求新增加的本质不同的回文串的数量

大致思路：

和 URAL 1940 Palindromes and Super Abilities一样就是简单的Palindromic Tree的应用, 就是时间有些紧, 所以读入和输出需要一些优化

代码如下：

Result  :  Accepted     Memory  :  88406 KB     Time  :  0.14 s

/*
* Author: Gatevin
* Created Time: 2015/3/30 20:52:20
* 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 5000010

char s[maxn];
char ans[maxn];
//string s, ans;
struct Palindromic_Tree
{
struct node
{
int next[2];
int len, sufflink;
};
node tree[maxn];
int L, len, suff;
void newnode()
{
L++;
for(int i = 0; i < 2; i++)
tree[L].next[i] = -1;
tree[L].len = tree[L].sufflink;
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;
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;
}
}
return true;
}
};

Palindromic_Tree pal;

int main()
{
ios_base::sync_with_stdio (false);//加上这句用cin, cout string 类型就不会TLE了
/*
cin>>s;
int n = s.length();
*/
/*
* 另外我用scanf("%s", s); printf("%s", ans)也是TLE
* 用puts和gets就没TLE了..
*/
gets(s);
int n = strlen(s);
pal.init();
for(int i = 0; i < n; i++)
ans[i] = '0' + pal.addLetter(i);
//cout<<ans;
puts(ans);
return 0;
}