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