Wikioi 1051 hash table

Description Description

N words are given and sorted by length. If a word I is the prefix of a word J, I-> J is counted as a solitaire (two identical words cannot be counted as a solitaire ).

Your task is to find the longest dragon for the input words.

Input description Input description

First Act N (1 <=n <= 105 ). Each of the following n rows contains one word (in lowercase), which is sorted by length. (Length of each word <50)

Output description Output description

The maximum length of a dragon.

Sample Input Sample Input

5

I

A

Int

Able

Inter

Sample output Sample output

3

Data range and prompt Data size & hint

1 <= n <= 105

Train of Thought: This question is whether to use the hash function to match the string or not, but it seems wise to be a stack application in terms of processing, because I didn't quite understand this question at the beginning, so I paid 100 points to download others.
It takes a long time to understand the code. In the sorting of Two-Dimensional String arrays, the sort function is further deepened and cannot be directly sorted. The two-dimensional arrays cannot be directly sorted in sort. Stack raises the time
It is much higher than linear conditions.

#include <cstdio>#include <cstring>#include <algorithm>#include<iostream>using namespace std;struct node{    int len;    char a[55];}e[100005];int Stack[100005];bool operator < (node b,node c){    for(int i=0;i<c.len&&i<b.len;i++)    {        if(b.a[i]<c.a[i]) return true;        else if(b.a[i]>c.a[i]) return false;    }    return b.len<c.len;}int hash(char *s,int len){    int  sum=0;    for(int i=0;i<len;i++)        sum=sum*26+s[i]-'a';    return sum;}int main(){    int n,i,sum=0,top=0;    cin>>n;    for(i=1; i<=n; i++)    {        scanf("%s",e[i].a);        e[i].len=strlen(e[i].a);    }    sort(e+1,e+n+1);    for(i=1;i<=n;i++)    {        while((top)&&e[i].len<=e[Stack[top]].len||hash(e[Stack[top]].a,e[Stack[top]].len)!=hash(e[i].a,e[Stack[top]].len))            top--;        Stack[++top]=i;        if(sum<top) sum=top;        //cout<<sum<<' '<<top<<' '<<i<<endl;    }    cout<<sum<<endl;    return 0;}