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）

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

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cmath>
6 #include<math.h>
7 #include<algorithm>
8 #include<queue>
9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define max(a,b) (a) > (b) ? (a) : (b)
18 #define min(a,b) (a) < (b) ? (a) : (b)
19 #define ll long long
20 #define eps 1e-10
21 #define MOD 1000000007
22 #define N 506
23 #define M 2006
24 #define inf 1e12
25 int n;
26 char s[N][M];
27 int vis[N];
28 int main()
29 {
30    int t;
31    int ac=0;
32    scanf("%d",&t);
33    while(t--){
34       memset(vis,0,sizeof(vis));
35       scanf("%d",&n);
36       int ans=-1;
37       for(int i=1;i<=n;i++){
38          scanf("%s",s[i]);
39          for(int j=i-1;j>=1;j--){
40             if(vis[j]) continue;
41             if(strstr(s[i],s[j])==0) ans=i;
42             else vis[j]=1;
43          }
44       }
45
46       printf("Case #%d: ",++ac);
47
48       if(ans==-1){
49          printf("-1\n");
50       }else{
51          printf("%d\n",ans);
52       }
53    }
54     return 0;
55 }
View Code