[hdu6230]Palindrome

Time Limit: 6000/3000 MS (Java/Others)
Memory Limit: 262144/262144 K (Java/Others)

Problem Description
Alice like strings, especially long strings. For each string, she has a special evaluation system to judge how elegant the string is. She defines that a string $S[1..3n−2](n≥2)$ is one-and-half palindromic if and only if it satisfies $S[i]=S[2n−i]=S[2n+i−2](1≤i≤n)$.For example, $abcbabc$ is one-and-half palindromic string, and $abccbaabc$ is not. Now, Alice has generated some long strings. She ask for your help to find how many substrings which is one-and-half palindromic.

Input
The first line is the number of test cases. For each test case, there is only one line containing a string(the length of strings is less than or equal to $500000$), this string only consists of lowercase letters.

Output
For each test case, output a integer donating the number of one-and-half palindromic substrings.

Sample Input

1
ababcbabccbaabc

Sample Output

2

Hint

In the example input, there are two substrings which are one-and-half palindromic strings, $abab$ and $abcbabc$.

题意：
输出字符串s中，"一又一半字符串"的数量
“一又一半字符串” : 字符串 $S[1..3n−2](n≥2)$，满足了 $S[i]=S[2n−i]=S[2n+i−2](1≤i≤n)$

题解：
很显然这个字符串是一个奇数长度的字符串。
我们先用马拉车做出对于每个中点 $i$的最长回文串长度的半径 $r[i]$。
然后很显然，我们考虑两个拼在一起的回文串的中点 $i,j(i>j)$
只需要满足 $j+r[j]-1<=i$且 $i-r[i]+1>=j$
那么我们把所有 $i-r[i]+1$相同的 $i$存在一起
然后枚举 $j=i-r[i]+1$，把所有的这些 $i$的树状数组位置都+1
然后查询区间 $[j,j+r[j]-1]$区间的元素之和。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline int lowbit(int x){return x&(-x);}
char s[500004];
int n,r[500004];
int tr[500004];
vector<int>vec[500004];
ll ans;
void add(int x){
    for(int i=x;i<=n;i+=lowbit(i))tr[i]++;
}
int ask(int x){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i))ret+=tr[i];
    return ret;
}
int w33ha(){
    ans=0;
    scanf("%s",s);
    n=strlen(s);
    for(int i=0;i<=n;i++)tr[i]=0;
    for(int i=0;i<=n;i++)vec[i].clear();
    for(int i=0;i<n;i++)r[i]=0;
    int mx=0,id=0;
    for(int i=0;i<n;i++){
        r[i]=(mx>i)?min(r[id*2-i],mx-i):1;
        while(i-r[i]>=0&&i+r[i]<n&&s[i-r[i]]==s[i+r[i]])r[i]++;
        if(i+r[i]>mx){
            mx=i+r[i];
            id=i;
        }
    }
    for(int i=n;i>=1;i--)r[i]=r[i-1];
    for(int i=1;i<=n;i++){
        vec[i-r[i]+1].push_back(i);
    }
    for(int i=1;i<=n;i++){
        for(int j=0;j<vec[i].size();j++){
            add(vec[i][j]);
        }
        ans+=ask(i+r[i]-1)-ask(i);
    }
    printf("%lld\n",ans);
    return 0;
}
int main(){
    int T;scanf("%d",&T);
    while(T--)w33ha();
    return 0;
}