HDU 5510：Bazinga（暴力KMP）

http://acm.hdu.edu.cn/showproblem.php?pid=5510

Bazinga

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：给出n个串，求出最大的i使得有一个j（1<=j<i）不是i的子串。

思路：比赛的时候由于没剩下多少时间，写的太暴力了TLE，后面自己写了一个，发现别人写的也很暴力，还有用strstr过的，速度普遍比用KMP的快。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define M 2010
 #define N 510
 int nxt[M];
 char s[N][M];
 
 void make_next(char *s)
 {
     memset(nxt, , sizeof(nxt));
     int j = -, i = ;
     nxt[] = -;
     int len = strlen(s);
     while(i < len) {
         if(j == - || s[i] == s[j]) {
             i++; j++;
             nxt[i] = j;
         } else {
             j = nxt[j];
         }
     }
 }
 
 int kmp(char *s, char *str)
 {
     make_next(s);
     int l = strlen(s), len = strlen(str);
     int i = , j = ;
     while(i < len && j < l) {
         if(j == - || str[i] == s[j]) {
             i++; j++;
             if(j >= l) return true;
         } else {
             j = nxt[j];
         }
     }
     return false;
 }
 
 int main()
 {
     int t;
     int cas = ;
     scanf("%d", &t);
     while(t--) {
         int n;
         scanf("%d", &n);
         for(int i = ; i <= n; i++)
             scanf("%s", s[i]);
 
         int tmp = -;
         for(int i = n; i > ; i--){
             if(!kmp(s[i-], s[i])) {
                  tmp = i;
                  break;
             }
         }
 //        printf("TMP : %d\n", tmp);
         int ans = tmp;
         for(int i = tmp - ; i > ; i--) {
             while(!kmp(s[i], s[ans+]) && ans < n) {
                 ans++;
             }
         }
 
         printf("Case #%d: ", ++cas);
         if(tmp != -)
             printf("%d\n", ans);
         else
             printf("-1\n");
     }
     return ;
 }