HDU_5510_Bazinga

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5056    Accepted Submission(s): 1596

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）

 

• 找到最大ID字符串i使得存在j<i的字符串不是i的子串
• 首先考虑如果从l到r串都满足都是r的子串，那么k>r串只需要检测r串即可
• 如果自l串开始到l+k串都是存在小于l的串不是当前串子串，那么对于l+k+1串就要检测l-1串和l到l+k串
• 算法出来了，就是记录当前最大id使得此id之前全是id串的子串，然后检测串的范围就是这个id到当前ID的前一位
• 如果用这样的算法，用c的strstr即可，用kmp可以减少一半时间

 #include <iostream>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <climits>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long           LL ;
 typedef unsigned long long ULL ;
 const int    maxn = 1e5 +    ;
 const int    inf  = 0x3f3f3f3f ;
 const int    npos = -         ;
 const int    mod  = 1e9 +     ;
 const int    mxx  =  +     ;
 const double eps  = 1e-       ;
 const double PI   = acos(-1.0) ;
 
 int T, n, use[maxn], mx, ans;
 char s[][];
 std::vector<int> v;
 int main(){
     // freopen("in.txt","r",stdin);
     // freopen("out.txt","w",stdout);
     while(~scanf("%d",&T)){
         for(int kase=;kase<=T;kase++){
             ans=-;
             scanf("%d",&n);
             mx=;
             use[]=;
             v.clear();
             scanf("%s",s[]);
             for(int i=;i<=n;i++){
                 scanf("%s",s[i]);
                 int flag=;
                 if(strstr(s[i],s[mx])==NULL)
                     flag=;
                 if(flag&&use[i-]){
                     for(int j=v.size()-;j>= && flag;j--){
                         if(strstr(s[i],s[v[j]])==NULL)
                             flag=;
                     }
                 }
                 if(flag){
                     mx=i;    use[i]=;    v.clear();
                 }else{
                     ans=i;    use[i]=;    v.push_back(i);
                 }
             }
 
             printf("Case #%d: %d\n",kase,ans);
         }
     }
     return ;
 }