hust 1400 Longest common prefix

题目描述

       Given N string S[1], S[2]......, S[N], you are to find a pair of strings whose longest common prefix has the largest length.

输入

       There are multiple test cases. For each case:

       The first line is a integer N. (2 <= N <= 20000)

       The next N lines each contains a non-empty string S[i]. The length of S[i] is at most 50.

输出

       For each case output a line representing the length of longest common prefix.

样例输入

3
bcabc
bcd
abcd

样例输出

2

我只想说这个题真的不难，可是它是我提交次数最多的几题了，我真的晕死了

这是一道Trie树的应用题，在每次插入的时候记录一下就可以了，可是我一开始用链表写，不知道内存超出了多少次，后来改用数组，还是超内存，不知道为什么，后

仔细的看了看这个题，发现不必弄那么多变量，于是就把所有变量去掉，就是

1. structTrie
2. {
3. bool isStr;
4. int step;
5. intnext;
6. }trie[maxn][26];

改成

1. int trie[maxn][26];

这样这个题才过，而且运行错误无数次

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f
 
using namespace std;
 
const double pi=acos(-1.0);
const double eps=1e-8;
typedef pair<int,int>pii;
 
const int maxn=1000000;
 
 
int trie[maxn][26];
 
int main()
{
    //freopen("in.txt","r",stdin);
    int cur,len;
    int alloc=1,n,m,sum;
    char s[51];
 
    while(scanf("%d",&n)!=EOF)
    {
 
    memset(trie,-1,sizeof(trie));
    sum=0;
    alloc=1;
    for (int i=0;i<n;i++)
    {
        scanf("%s",s);
        len=strlen(s);
        cur=0;
        for (int j=0;j<len;j++)
        {
            if (trie[cur][s[j]-'a']==-1)
            {
                trie[cur][s[j]-'a']=alloc;
                //trie[cur][s[j]-'a'].step=j+1;
                //sum=max(trie[cur][s[j]-'a'].step,sum);
                alloc++;
            }
            else sum=max(j+1,sum);
            //if (j==len-1) trie[cur][s[j]-'a'].isStr=true;
            cur=trie[cur][s[j]-'a'];
        }
    }
    printf("%d
",sum);
    }
    //fclose(stdin);
    return 0;
}