HDU5658 CA Loves Palindromic(回文树)

描述

传送门：我是传送门

CA loves strings, especially loves the palindrome strings. One day he gets a string, he wants to know how many palindromic substrings in the substring S[l,r]S[l,r]. Attantion, each same palindromic substring can only be counted once.

输入

First line contains TT denoting the number of testcases. TT testcases follow. For each testcase: First line contains a string SS. We ensure that it is contains only with lower case letters. Second line contains a interger QQ, denoting the number of queries. Then QQ lines follow, In each line there are two intergers l,rl,r, denoting the substring which is queried. 1≤T≤101≤T≤10, 1≤length≤1000,1≤Q≤100000,1≤l≤r≤length1≤length≤1000,1≤Q≤100000,1≤l≤r≤length

输出

For each testcase, output the answer in QQ lines.

样例

输入

1
abba
2
1 2
1 3

输出

2
3

Note

In first query, the palindromic substrings in the substring S[1,2]S[1,2] are “a”,”b”.
In second query, the palindromic substrings in the substring S[1,2]S[1,2] are “a”,”b”,”bb”.
Note that the substring “b” appears twice, but only be counted once.
You may need an input-output optimization.

思路

利用回文树的性质暴力预处理所有子串

每次查询都是0(1)0(1)的复杂度

不同回文子串的数量=P−2不同回文子串的数量=P−2

代码

/*
 * =========================================================================
 *
 *       Filename:  C.cpp
 *
 *           Link:  http://acm.hdu.edu.cn/showproblem.php?pid=5658
 *
 *        Version:  1.0
 *        Created:  2018/09/03 04时11分41秒
 *       Revision:  none
 *       Compiler:  g++
 *
 *         Author:  杜宁元 (https://duny31030.top/), duny31030@126.com
 *   Organization:  QLU_浪在ACM
 *
 * =========================================================================
 */
#include <bits/stdc++.h>
using namespace std;
#define clr(a, x) memset(a, x, sizeof(a))
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define pre(i,a,n) for(int i=a;i>=n;i--)
#define ll long long
#define max3(a,b,c) fmax(a,fmax(b,c))
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int MAXN = 1e5+10;
const int N = 26;
char s[MAXN];
struct node
{
    int next[N];   // next指针,next指针和字典树类似,指向的串为
                   // 当前穿两端加上同一个字符构成
    int fail;      // fail指针,失配后跳转到fail指针指向的节点
    int cnt;       // 表示节点i的本质不同的串的个数
                   // 建树时求出的不是完全的，最后Count()函数跑一遍以后才是正确的
    int num;       // 表示以节点i表示的最长回文串的最右端点为回文串结尾的回文个数
    int len;       // 表示以当前节点表示的最长回文串长度（一个节点表示一个回文串）
    int S;         // 存放添加的字符
}pam[MAXN];
int last;   // 指向新添加一个字母后所形成的最长回文串表示的节点
int n;      // 表示添加的字符个数
int p;      // 表示添加的节点个数

int newnode(int x)
{
    for(int i = 0;i < N;i++)
        pam[p].next[i] = 0;
    pam[p].cnt = 0;
    pam[p].num = 0;
    pam[p].len = x;
    return p++;
}

void Init()
{
    p = 0;
    newnode(0);
    newnode(-1);
    last = 0;
    n = 0;
    pam[n].S = -1;   // 开头存放一个字符集中没有的字符,减少特判
    pam[0].fail = 1;
}

int get_fail(int x)
{
    while(pam[n-pam[x].len-1].S != pam[n].S)
        x = pam[x].fail;
    return x;
}

void Add(int c)
{
    c -= 'a';
    pam[++n].S = c;
    int cur = get_fail(last);   // 通过上一个回文串找这个回文串的匹配位置
    if(!pam[cur].next[c])       // 如果这个回文串没有出现过，说明出现了
                                // 一个新的本子不同的回文串
    {
        int now = newnode(pam[cur].len+2);   // 新建节点
        pam[now].fail = pam[get_fail(pam[cur].fail)].next[c];   
        // 和AC自动机一样建立fil指针,以便失败后回跳
        pam[cur].next[c] = now;
        pam[now].num = pam[pam[now].fail].num+1;
    }
    last = pam[cur].next[c];
    pam[last].cnt++;
}

void Count()
{
    for(int i = p-1;i >= 0;--i)
    {
        pam[pam[i].fail].cnt += pam[i].cnt;
        // 父亲累加儿子才cnt,因为如果fail[v] = u , 则u一定是v的子回文串
    }
}
int T,que,l,r;
int main()
{
    ios
#ifdef ONLINE_JUDGE 
#else 
        freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout); 
#endif
    int pr[1010][1010];
    memset(pr,0,sizeof pr);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        int len = strlen(s);
        for(int i = 0;i < len;i++)
        {
            Init();
            for(int j = i;j < len;j++)
            {
                Add(s[j]);
                pr[i][j] = p-2;
            }
        }
        scanf("%d",&que);
        while(que--)
        {
            scanf("%d %d",&l,&r);
            printf("%d
",pr[l-1][r-1]);
        }
    }
    fclose(stdin);
    // fclose(stdout);
    return 0;
}