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".
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.
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。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 #define M 2010 6 #define N 510 7 int nxt[M]; 8 char s[N][M]; 9 10 void make_next(char *s) 11 { 12 memset(nxt, 0, sizeof(nxt)); 13 int j = -1, i = 0; 14 nxt[0] = -1; 15 int len = strlen(s); 16 while(i < len) { 17 if(j == -1 || s[i] == s[j]) { 18 i++; j++; 19 nxt[i] = j; 20 } else { 21 j = nxt[j]; 22 } 23 } 24 } 25 26 int kmp(char *s, char *str) 27 { 28 make_next(s); 29 int l = strlen(s), len = strlen(str); 30 int i = 0, j = 0; 31 while(i < len && j < l) { 32 if(j == -1 || str[i] == s[j]) { 33 i++; j++; 34 if(j >= l) return true; 35 } else { 36 j = nxt[j]; 37 } 38 } 39 return false; 40 } 41 42 int main() 43 { 44 int t; 45 int cas = 0; 46 scanf("%d", &t); 47 while(t--) { 48 int n; 49 scanf("%d", &n); 50 for(int i = 1; i <= n; i++) 51 scanf("%s", s[i]); 52 53 int tmp = -100; 54 for(int i = n; i > 1; i--){ 55 if(!kmp(s[i-1], s[i])) { 56 tmp = i; 57 break; 58 } 59 } 60 // printf("TMP : %d ", tmp); 61 int ans = tmp; 62 for(int i = tmp - 1; i > 0; i--) { 63 while(!kmp(s[i], s[ans+1]) && ans < n) { 64 ans++; 65 } 66 } 67 68 printf("Case #%d: ", ++cas); 69 if(tmp != -100) 70 printf("%d ", ans); 71 else 72 printf("-1 "); 73 } 74 return 0; 75 }