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.
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： T组数据 每组数据给出n个字符串 要求找出第i个串（i要尽可能的大）不包含i之前的某个或多个字符串
分析： dfs写从最后一个字符串向前搜找到相邻的2个不互相包含的字符串，
#include<cstdio>
#include<cstring>
#include<stack>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int oo = 1e9+7;
const int maxn = 1e6+7;
char str[510][2010];
int n, ans;
void dfs(int x)
{
int i;
if(x <= 1 || ans == n) return ;
if(strstr(str[x], str[x-1])!=NULL)
dfs(x-1);
else
{
for(i = n; i > x-1; i--)///搜到当前位置
{
if(strstr(str[i], str[x-1]) == NULL)
{
ans = max(i, ans);
}
}
dfs(x-1);///可能还会存在前面的字符串符合条件 继续搜。
}
}
int main()
{
int i, T, cas=1;
scanf("%d", &T);
while(T--)
{
ans = -1;
scanf("%d", &n);
for(i = 1; i <= n; i++)
scanf("%s", str[i]);
dfs(n);
printf("Case #%d: %d
", cas++, ans);
}
return 0;
}