hdu--5510

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1472    Accepted Submission(s): 459

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

45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc

 

Sample Output

Case #1: 4Case #2: -1Case #3: 4Case #4: 3
解题思路：在写这题时需要用到匹配算法，一般想到节省时间的就是用kmp，但在c++中提供了一个函数strstr(str1,str2) 函数用于判断字符串str2是否是str1的子串。如果是，则该函数返回str2在str1中首次出现的地址；否则，返回NULL。所以用这个函数更为方便。这题需要剪枝否则会超时。剪枝：首先对每对相邻的子串进行判断，如果编号小的字符串是编号大的字符串的子串，那么该编号小的字符串在后续的判断中无需出现了。接着就从最后一个字符串开始扫描判断就OK了。
#include<stdio.h>#include<string.h>char s[510][2005];int num[510];int main(){int t,n,i,j,k;k=1;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++){scanf("%s",s[i]);}//输入 int v=-1;memset(num,0,sizeof(num));for(i=1;i<n;i++){if(strstr(s[i+1],s[i])!=NULL){num[i]=1;}}//剪枝       for(i=n;i>=2;i--){      for(j=i-1;j>=1;j--){      if(num[j])continue;      int l1=strlen(s[i]);      int l2=strlen(s[j]);      if(l1<l2)continue;      if(strstr(s[i],s[j])==NULL){      v=i;      break;      }      }      if(v!=-1)break;      }      printf("Case #%d: %d\n",k++,v);    }return 0;}