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

   
   
   
   

    
    
    
    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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    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;
}