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 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

 

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=0;
   scanf("%d",&t);
   while(t--){
      memset(vis,0,sizeof(vis));
      scanf("%d",&n);
      int ans=-1;
      for(int i=1;i<=n;i++){
         scanf("%s",s[i]);
         for(int j=i-1;j>=1;j--){
            if(vis[j]) continue;
            if(strstr(s[i],s[j])==0) ans=i;
            else vis[j]=1;
         }
      }
      
      printf("Case #%d: ",++ac);
      
      if(ans==-1){
         printf("-1\n");
      }else{
         printf("%d\n",ans);
      }
   }
    return 0;
}