Bazinga

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5510

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

HINT

题意

你需要找到一个最大的i使得，存在一个在他前面的字符串不是他的子串

题解：

就暴力匹配就好了，然后加一个剪枝，如果这个字符串是某个字符串的子串的话，就不用检查他了（讲道理的话，这个剪枝是没有用的，因为全部都不是子串的话，这个剪枝没有一点卵用。只是数据出水了而已……

正解应该是后缀自动机？AC自动机？

出题人的意思是只用检查相邻的两个字符串，好像很有道理的样子~

代码

#include<iostream>

#include<stdio.h>

#include<cstring>

using namespace std;

char s[][];

int vis[];

int main()

{

    int t;scanf("%d",&t);

    for(int cas=;cas<=t;cas++)

    {

        memset(vis,,sizeof(vis));

        int n;scanf("%d",&n);

        int flag = ;

        for(int i=;i<=n;i++)

        {

            scanf("%s",s[i]);

            for(int j=i-;j>=;j--)

            {

                if(vis[j])continue;

                if(strstr(s[i],s[j])==NULL)flag=i;

                else vis[j]=;

            }

        }

        if(!flag)

            printf("Case #%d: -1\n",cas);

        else

            printf("Case #%d: %d\n",cas,flag);

    }

}

巴特西

HDU 5510 Bazinga 暴力匹配加剪枝

Bazinga

题目连接

Description

HINT

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