Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2390    Accepted Submission(s): 746

 

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

题意：

求i最大的且其前无子串的字符串，若无则输出-1。

我们运用了求子串函数strstr();   关于strstr();的介绍请见：http://www.cnblogs.com/Kiven5197/p/5869909.html

先找两个相邻 若找出不匹配的串再反向对比。

附AC代码：

 1 #include<bits/stdc++.h> 2 using namespace std; 3  4 //strstr(); 5  6 char s[510][2010]; 7  8 int main(){ 9     int t,n;10     scanf("%d",&t);//cin>>t;11     int ans=1;12     while(t--){13         scanf("%d",&n);//cin>>n;14         for(int i=1;i<=n;i++){15             scanf("%s",s[i]);//cin>>s[i];16         }17         int res=-1;18         for(int i=n;i>0;i--){19             if(!strstr(s[i],s[i-1])){20                 res=max(i,res);21                 for(int j=i+1;j<=n;j++){22                     if(!strstr(s[j],s[i-1])){23                         res=max(j,res);24                     }25                 }26             }27         }28         printf("Case #%d: %d\n",ans++,res);//cout<<"Case #"<<ans++<<": "<<res<<endl;29     }30     return 0;31 }

这题第一发用的cin和coutT掉了..改成scanf printf就A了

HDU-5510 Bazinga