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

45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc

 

Sample Output

Case #1: 4Case #2: -1Case #3: 4Case #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;}