HDU 5510---Bazinga（指针模拟）

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.

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.

 

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

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5910 5909 5908 5907 5906 

 

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[600][2005];
int nex[2005];

void makeNext(char p[])
{
    int q,k;
    int m=strlen(p);
    nex[0]=0;
    for(q=1,k=0; q<m; ++q)
    {
        while(k>0&&p[q]!=p[k])
            k=nex[k-1];
        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=0,k=0; q<len; q++)
    {
        while(k>0&&s[y][q]!=s[x][k])
            k=nex[k-1];
        if(s[y][q]==s[x][k]) k++;
        if(k>=l) return 1;
    }
    return 0;
}

int main()
{
    int T,Case=1;
    cin>>T;
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%s",s[i]);
        int l = 1, r = 2, ans = -1;
        while(r <= n)
        {
            while(l < r)
            {
                if(calc(l,r))  l++;
                else { ans=r; break; }
            }
            r++;
        }
        printf("Case #%d: %d
",Case++,ans);
    }
    return 0;
}