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 2000letters. 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：给定n个串，求最大的下表i使得在1~i-1中至少有一个串不是其子串



思路： 暴力O(T*n*n*(n+m)) 于是TLE

优化~用l代表对于当前i，所能抵达的l。如果s[l]不是s[i]的字串，break。 如果是，l++;


#include <stdio.h>
#include <string.h>
typedef long long ll;

using namespace std;

const int N=2e3+5;
char s[501][N];
int next[N];
int lena,lenb;
char a[N];
char b[N];
ll ans;

void set_naxt()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<lenb)
    {
        if(j==-1||b[i]==b[j])
        {
            i++; j++;
            next[i]=j;
        }
        else
        j=next[j];
    }
}

int kmp()
{
    int i=0,j=0;
    set_naxt();
    while(i<lena)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;j++;
        }
        else
        j=next[j];
        if(j==lenb)
        {
            return 1;
        }
    }
    return 0;
}

int main(void){
    int T;
    int ks=1;
    scanf("%d",&T);
    while(T--){
        ans=-1;
        int m;
        scanf("%d",&m);
        register int i,l=1;
        bool flag=true;
        for(i=1;i<=m;i++)   scanf("%s",s[i]);
        for(i=1;i<=m;i++){
            while(l<i){
                strcpy(a,s[i]),strcpy(b,s[l]);
                lena=strlen(a),lenb=strlen(b);
                if(kmp()==1)    l++;
                else    {ans=i;break;}
            }
        }
        printf("Case #%d: %d\n",ks++,ans);
    }
}