fir n) can %s all 代碼 -m panel led

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". InputThe 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.OutputFor 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

思路：
暑訓的時候曾經寫過這道題，不過我竟然忘了正解，實際上這題還是比較暴力的，由於題目要求的只是哪一個存在就行啦，所以在層層嵌套的情況下，不需要完全直接掃一編。
於是，我們首先處理出，相鄰的有哪些是滿足字串的關系的，然後再做處理就行啦。
代碼

#include<iostream>
#include<cstring>
using namespace std;
char a[505][2048];
int num[600];
int main()
{
    int T;
    scanf("%d",&T);
    int Ca = 0;
    while(T--){
        Ca++;
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%s",a[i]);
        }
        int ans=0;
        for(int i=1;i<n;i++){
            num[i]=strstr(a[i+1],a[i])?1:0;
        }
        for(int i=n;i>=1;i--){
            for(int j=1;j<i;j++){
                if(num[j]){continue;}
                if(!strstr(a[i],a[j])){
                    ans=i;
                    break;
                }
            }
            if(ans){break;}
        }
        printf("Case #%d: ",Ca);
        if(ans){printf("%d\n",ans);}
        else printf("-1\n");
    }
}

HDU - 5510 Bazinga