• CF245H Queries for Number of Palindromes(回文树)


    题意翻译

    题目描述

    给你一个字符串s由小写字母组成,有q组询问,每组询问给你两个数,l和r,问在字符串区间l到r的字串中,包含多少回文串。

    输入格式

    第1行,给出s,s的长度小于5000 第2行给出q(1<=q<=10^6) 第2至2+q行 给出每组询问的l和r

    输出格式

    输出每组询问所问的数量。

    题目描述

    You've got a string s=(s_{1})(s_{2})...(s_{|s|}) of length |s| , consisting of lowercase English letters. There also are qq queries, each query is described by two integers (l_{i}),(r_{i}) (1<=(l_{i})<=(r_{i})<=|s|) . The answer to the query is the number of substrings of string (s[) (l_{i}) ... (r_{i}) (]) , which are palindromes.

    String (s[l... r])=(s_{l})(s_{l+1})...( s_{r})(1<=l<=r<=∣s∣) is a substring of string (s=s_{1}s_{2}... s_{|s|}) .

    String tt is called a palindrome, if it reads the same from left to right and from right to left. Formally, if (t=t_{1}t_{2}... t_{|t|}=t_{|t|}t_{|t|-1}... t_{1}).

    输入输出格式

    输入格式:

    The first line contains string ss (1<=|s|<=5000) . The second line contains a single integer qq (1<=q<=106) — the number of queries. Next qq lines contain the queries. The ii -th of these lines contains two space-separated integers (l_{i}),(r_{i}) (1<=(l_{i})<=(r_{i})<=|s|) — the description of the i-th query.

    It is guaranteed that the given string consists only of lowercase English letters.

    输出格式:

    Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

    输入输出样例

    输入样例#1: 复制

    caaaba
    5
    1 1
    1 4
    2 3
    4 6
    4 5

    输出样例#1: 复制

    1
    7
    3
    4
    2

    说明

    Consider the fourth query in the first test case. String (s[4... 6]) = «aba». Its palindrome substrings are: «a», «b», «a», «aba».


    题解

    这个题目的思路非常巧妙?
    因为时间复杂度允许达到(n)2,于是我们就从1开始一直到strlen(s),(l)的位置每向后移动一位就清空回文树,并把这个(l...len)的回文串重新放入回文树。用(ans[l][r])来统计一下答案
    然后o(1)查询就OK了。
    然后我yy了一下莫队?
    是不是离线的话时间复杂度就降到了(n{sqrt n})了呢?


    代码

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    struct node{
        int fail,len,ch[26],dep;
    }t[5001];
    int tot,k,ans[5001][5001];
    char s[5001],ch[5001];
    int read()
    {	
        int x=0,w=1;char ch=getchar();
        while(ch>'9'||ch<'0'){if(ch=='-')w=-1;ch=getchar();}
        while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
        return x*w;
    }
    
    void clear()
    {
        memset(t,0,sizeof(t));
        tot=1;k=0;t[0].fail=t[1].fail=1;t[1].len=-1;
    }
    
    void solve()
    {
        int n=strlen(ch+1);
        for(int i=1;i<=n;i++)
        {
            clear();
            for(int j=i;j<=n;j++)s[j-i+1]=ch[j];
            for(int j=i;j<=n;j++)
            {
                //clear();
                while(s[j-i+1-t[k].len-1]!=s[j-i+1])k=t[k].fail;
                if(!t[k].ch[s[j-i+1]-'a']){
                    t[++tot].len=t[k].len+2;
                    int l=t[k].fail;
                    while(s[j-i+1-t[l].len-1]!=s[j-i+1])l=t[l].fail;
                    t[tot].fail=t[l].ch[s[j-i+1]-'a'];
                    t[k].ch[s[j-i+1]-'a']=tot;
                    t[tot].dep=t[t[tot].fail].dep+1;
            }
                k=t[k].ch[s[j-i+1]-'a'];
                ans[i][j]=ans[i][j-1]+t[k].dep;
            }
        }
    }
    
    int main()
    {
        scanf("%s",ch+1);
        solve();
        int q=read();
        for(int i=1;i<=q;i++)
        {	
        int l=read(),r=read();
        printf("%d
    ",ans[l][r]);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/hhh1109/p/9246094.html
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