Problem 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：给n个字符串，编号1到n，找最大的编号j，使得存在1到 j-1 的某个串非 j 的字串

思路：kmp判断是否是字串，从小到大遍历，用i，j指针

#include<stdio.h>
#include<vector>
#include<queue>
#include <cstring>
using namespace std;

#define move(a) (1<<(a))
const int N = 1010;
const int inf = 0x3fffffff;

char s[510][2100];
int nextt[2100];

int kmp(char *s1,char *s2){
    int i,j,len1,len2;
    len1=strlen(s1);
    len2=strlen(s2);
    i=0;j=-1;
    memset(nextt,-1,sizeof(nextt));
    while(i<len2){
        if(j==-1 || s2[i]==s2[j]){
            i++;
            j++;
            nextt[i]=j;
        }
        else j=nextt[j];
    }
    i=0;j=0;
    while(i<len1 && j<len2){
        if(j==-1 || s1[i]==s2[j]){
            i++;
            j++;
        }
        else j=nextt[j];
    }
    if(j>=len2)
    return 1;
    else return 0;
}

int T,n;
int main()
{
    int cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",&s[i]);
        }

        int ans=-1;
        int i=1,j=2;
        while(i<=n && j<=n)
        {
            while(i<j&&kmp(s[j],s[i])) i++;
            if(i<j) ans=j;
            j++;

        }

        printf("Case #%d: %d\n",cas++,ans);
    }
    return 0;
}

TLE的代码：（思路是 比较上有点繁琐）

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
typedef  long long LL ;

const LL mod=1000000009LL;
const int N=10000;
const int inf=0x3fffffff;

int T;
int n,m,k;
char s[2100];

char sta[510][2100];
int Next[2010][510];
int slen[510];
int tot=0;

void MakeNext(char p[],int len,int tot)
{
    Next[0][tot]=-1;
    int i=0;
    int j=-1;
    while(i<len)
    {
        if(j==-1||p[i]==p[j])
        {
            i++,j++;
            if(p[i]!=p[j]) Next[i][tot]=j;
            else Next[i][tot]=Next[j][tot];
        }else
        {
            j=Next[j][tot];
        }
    }
}

int KMP(char T[],int N,char P[],int M,int tot)
{
    int i=0,j=0;
    while(i<N&&j<M)
    {
        if(T[i]==P[j]||j==-1) i++,j++;
        else j=Next[j][tot];
    }
    if(j==M) return i-M+1;
    else return -1;
}

int main()
{
    int cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        scanf("%s",sta[0]);

        tot=0;
        slen[0] = strlen(sta[0]);
        MakeNext(sta[0],slen[0],tot);
        tot++;

        int ans=-1;
        for(int t=1; t<n; t++)
        {
            scanf("%s",&s);
            int len = strlen(s);

            int old = tot,i;
            for(i=0; i<old; i++)
            {
                if(KMP(s,len,sta[i],slen[i],i)==-1)
                {
                    MakeNext(s,len,tot);
                    for(int j=0;j<len;j++)
                        sta[tot][j]= s[j];
                    slen[tot++]=len;
                    ans=t+1;
                    break;
                }
            }
            if(i==old)
            {
                tot=0;
                MakeNext(s,len,tot);
                for(int j=0;j<len;j++)
                    sta[tot][j]= s[j];
                slen[tot++]=len;
            }

//            for(int i=0;i<tot;i++)
//            {
//                printf("    %s\n",sta[i]);
//            }

        }

        printf("Case #%d: %d\n",cas++,ans);
    }
    return 0;
}