bazinga
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
题解：
查找前串不是其子串的串，输出其序号，若没有，输出-1。因为串比较大，用kmp算法比较好。
代码如下：
#include<algorithm>
#include<iOStream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<queue>
#include<map>
#include<set>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define maxn 100007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1001113
//#define e 2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) prllf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
//const int maxn=;
using namespace std;
char a[507][2007];
int b[507];
int kmp(int p,int q)
{
int l=strlen(a[p]),ls=strlen(a[q]);
int i=0,j=0;
while(i<l&&j<ls)
{
if(a[p][i]==a[q][j])
{
i++;
j++;
}
else
{
i=i-j+1;
j=0;
}
}
if(j==ls)
return i-j;
return -1;
}
int main()
{
int t,s=1;
cin>>t;
while(t--)
{
int n;
cin>>n;
int k=0;
memset(b,0);
for(int i=1;i<=n;i++)
{
cin>>a[i];
for(int j=i-1; j>=1; j--)
{
if(b[j]!=0)
continue;
if(kmp(i,j)==-1)
{
k=i;
break;
}
else
b[j]=1;
}
}
if(!k)
cout<<"Case #"<<s++<<": -1"<<endl;
else
cout<<"Case #"<<s++<<": "<<k<<endl;
}
}