Bazinga

Time Limit: 20 Sec

Memory Limit: 256 MB

題目連接

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

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

HINT

?

題意

你需要找到一個(gè)最大的i使得，存在一個(gè)在他前面的字符串不是他的子串

題解：

就暴力匹配就好了，然后加一個(gè)剪枝，如果這個(gè)字符串是某個(gè)字符串的子串的話，就不用檢查他了（講道理的話，這個(gè)剪枝是沒有用的，因?yàn)槿慷疾皇亲哟脑?#xff0c;這個(gè)剪枝沒有一點(diǎn)卵用。只是數(shù)據(jù)出水了而已……

正解應(yīng)該是后綴自動(dòng)機(jī)？AC自動(dòng)機(jī)？

出題人的意思是只用檢查相鄰的兩個(gè)字符串，好像很有道理的樣子~

代碼

?

#include<iostream> #include<stdio.h> #include<cstring> using namespace std;char s[550][2005]; int vis[550]; int main() {int t;scanf("%d",&t);for(int cas=1;cas<=t;cas++){memset(vis,0,sizeof(vis));int n;scanf("%d",&n);int flag = 0;for(int i=1;i<=n;i++){scanf("%s",s[i]);for(int j=i-1;j>=1;j--){if(vis[j])continue;if(strstr(s[i],s[j])==NULL)flag=i;else vis[j]=1;}}if(!flag)printf("Case #%d: -1\n",cas);elseprintf("Case #%d: %d\n",cas,flag);} }

?

轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/4929963.html

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