题目链接
http://acm.hdu.edu.cn/search.php?action=listproblem
Problem Description
Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.
For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.
Don't tilt your head. I'm serious.
For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.
A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
Input
The first line contains an integer t (1≤t≤50) which is the number of test cases.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
Output
For each test case, output the largest label you get. If it does not exist, output −1.
Sample Input
4
5
ab
abc
zabc
abcd
zabcd
4
you
lovinyou
aboutlovinyou
allaboutlovinyou
5
de
def
abcd
abcde
abcdef
3
a
ba
ccc
Sample Output
Case #1: 4
Case #2: -1
Case #3: 4
Case #4: 3
Source
Recommend
题意：输入n 然后输入n个字符串，求最大的i 要求1~i-1中有一个串不是i的子串？
思路：分析复杂度可知这n个字符串比较次数只能是O（n） 那么发现可以用指针模拟的办法解决。具体做法：定义l和r 如果l是r的子串，那么l++ 继续判断l是否是r的子串，否则ans=r， 为什么这样呢？ 如果l是r的子串那么l++ 由它可知1~l-1 这些串一定是l~r这些串中一些串的子串，那么1~l-1这些串不必再和r后面的串进行比较；
代码如下：
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstring>
#include <queue> using namespace std;
typedef long long LL;
char s[][];
int nex[]; void makeNext(char p[])
{
int q,k;
int m=strlen(p);
nex[]=;
for(q=,k=; q<m; ++q)
{
while(k>&&p[q]!=p[k])
k=nex[k-];
if(p[q]==p[k]) k++;
nex[q]=k;
}
}
int calc(int x,int y)
{
makeNext(s[x]);
int l=strlen(s[x]);
int len=strlen(s[y]);
int q,k;
for(q=,k=; q<len; q++)
{
while(k>&&s[y][q]!=s[x][k])
k=nex[k-];
if(s[y][q]==s[x][k]) k++;
if(k>=l) return ;
}
return ;
} int main()
{
int T,Case=;
cin>>T;
while(T--)
{
int n;
scanf("%d",&n);
for(int i=; i<=n; i++)
scanf("%s",s[i]);
int l = , r = , ans = -;
while(r <= n)
{
while(l < r)
{
if(calc(l,r)) l++;
else { ans=r; break; }
}
r++;
}
printf("Case #%d: %d\n",Case++,ans);
}
return ;
}