HDU-5510 Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2390    Accepted Submission(s): 746

 

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 (1in) such that there exists an integer j (1j<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 (1t50) which is the number of test cases.
For each test case, the first line is the positive integer n (1n500) 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

 

题意:

求i最大的且其前无子串的字符串,若无则输出-1。

 

我们运用了求子串函数strstr();   关于strstr();的介绍请见:http://www.cnblogs.com/Kiven5197/p/5869909.html

先找两个相邻 若找出不匹配的串再反向对比。

 

附AC代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 //strstr();
 5 
 6 char s[510][2010];
 7 
 8 int main(){
 9     int t,n;
10     scanf("%d",&t);//cin>>t;
11     int ans=1;
12     while(t--){
13         scanf("%d",&n);//cin>>n;
14         for(int i=1;i<=n;i++){
15             scanf("%s",s[i]);//cin>>s[i];
16         }
17         int res=-1;
18         for(int i=n;i>0;i--){
19             if(!strstr(s[i],s[i-1])){
20                 res=max(i,res);
21                 for(int j=i+1;j<=n;j++){
22                     if(!strstr(s[j],s[i-1])){
23                         res=max(j,res);
24                     }
25                 }
26             }
27         }
28         printf("Case #%d: %d\n",ans++,res);//cout<<"Case #"<<ans++<<": "<<res<<endl;
29     }
30     return 0;
31 }

 

这题第一发用的cin和coutT掉了..改成scanf printf就A了

posted @ 2016-09-13 21:44  Kiven#5197  阅读(205)  评论(0编辑  收藏  举报