hdu5510 Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 532    Accepted Submission(s): 220

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

 

这是一道双指针kmp题，因为如果A是B的子串，那么后面判断的时候A就可以省略不判断，因为如果A不是其子串，B肯定不是其子串。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
char s[505][2005],nextt[2005];

int pd(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 main()
{
    int T,n,m,i,j,ans,num,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ans=-1;
        for(i=1;i<=n;i++){
            scanf("%s",s[i]);
        }
        j=1;
        for(i=2;i<=n;i++){     //这里用两个指针，i指当前位置，j指遍历到哪个串
            while(j<i && pd(s[i],s[j]) )j++;
            if(j<i)ans=i;
        }
        cas++;
        printf("Case #%d: %d\n",cas,ans);
    }
    return 0;
}