hdu 5510 Bazinga（暴力）

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  to n, you should find the largest i (≤i≤n) such that there exists an integer j (≤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 (≤t≤) which is the number of test cases.
For each test case, the first line is the positive integer n (≤n≤) 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  letters.

 

Output

For each test case, output the largest label you get. If it does not exist, output −.

 

Sample Input

ab
abc
zabc
abcd
zabcd

you
lovinyou
aboutlovinyou
allaboutlovinyou

de
def
abcd
abcde
abcdef

a
ba
ccc

Sample Output

Case #:
Case #: -
Case #:
Case #: 

 

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

直接暴力也能过，这是区域赛吗？

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 506
 #define M 2006
 #define inf 1e12
 int n;
 char s[N][M];
 int vis[N];
 int main()
 {
    int t;
    int ac=;
    scanf("%d",&t);
    while(t--){
       memset(vis,,sizeof(vis));
       scanf("%d",&n);
       int ans=-;
       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])==) ans=i;
             else vis[j]=;
          }
       }

       printf("Case #%d: ",++ac);

       if(ans==-){
          printf("-1\n");
       }else{
          printf("%d\n",ans);
       }
    }
     return ;
 }