2015ACM/ICPC亚洲区沈阳站 B-Bazinga

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3509    Accepted Submission(s): 1122

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

 

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

 

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/*
用两个指针模拟kmp，暴力
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define N 505
using namespace std;
///4294967295
int t,n;
int ne[2005];
void getnext(char *s2)
{
    int i=0,j=-1,l=strlen(s2);
    ne[0]=-1;
    while(i<l)
    {
        if(j==-1||s2[i]==s2[j])
        {
            i++;j++;
            if(s2[i]==s2[j])
                ne[i]=ne[j];
            else
                ne[i]=j;
        }
        else
            j=ne[j];
    }
}
int kmp(char *s,char *s2)
{
    memset(ne,0,sizeof(ne));
    getnext(s2);
    int i=0,j=0,k=0,len=strlen(s),l=strlen(s2);
    while(i<len)
    {
        if(j==-1||s[i]==s2[j])
            i++,j++;
        else
            j=ne[j];
        if(j==l)
            k++,j=ne[j];
    }
    return k;
}
char s[N][2005]={"000"};
bool vis[N],f;
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    scanf("%d",&t);
    int Case=1;
    while(t--)
    {
        memset(vis,true,sizeof vis);
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%s",s[i]);
        printf("Case #%d: ",Case++);
        for(int i=2;i<=n;i++)
            if(kmp(s[i],s[i-1])>0)
                vis[i-1]=false;
        int f=0;
        for(int i=n;i>=1&&!f;i--)
            for(int j=1;j<i&&!f;j++)
                if(vis[j])
                if(kmp(s[i],s[j])==0)
                {
                    printf("%d\n",i);
                    f=true;
                }
        if(!f)
           puts("-1");
    }
    return 0;
}