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HDU 5510---Bazinga（指针模拟）

2024-08-15 15:30:26   246人阅读

题目链接

http://acm.hdu.edu.cn/search.php?action=listproblem

 

Problem Description
Ladies and gentlemen, please sit up straight.
Don‘t tilt your head. I‘m serious.
技术分享

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
 
Source
2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）
 

 

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题意：输入n 然后输入n个字符串，求最大的i 要求1~i-1中有一个串不是i的子串？
 
思路：分析复杂度可知这n个字符串比较次数只能是O（n） 那么发现可以用指针模拟的办法解决。具体做法：定义l和r 如果l是r的子串，那么l++ 继续判断l是否是r的子串，否则ans=r， 为什么这样呢？  如果l是r的子串那么l++  由它可知1~l-1 这些串一定是l~r这些串中一些串的子串，那么1~l-1这些串不必再和r后面的串进行比较；
 
代码如下：
#include <iostream>#include <algorithm>#include <stdio.h>#include <cstring>#include <queue>using namespace std;typedef long long LL;char s[600][2005];int nex[2005];void makeNext(char p[]){    int q,k;    int m=strlen(p);    nex[0]=0;    for(q=1,k=0; q<m; ++q)    {        while(k>0&&p[q]!=p[k])            k=nex[k-1];        if(p[q]==p[k]) k++;        nex[q]=k;    }}int calc(int x,int y){    makeNext(s[x]);    int l=strlen(s[x]);    int len=strlen(s[y]);    int q,k;    for(q=0,k=0; q<len; q++)    {        while(k>0&&s[y][q]!=s[x][k])            k=nex[k-1];        if(s[y][q]==s[x][k]) k++;        if(k>=l) return 1;    }    return 0;}int main(){    int T,Case=1;    cin>>T;    while(T--)    {        int n;        scanf("%d",&n);        for(int i=1; i<=n; i++)            scanf("%s",s[i]);        int l = 1, r = 2, ans = -1;        while(r <= n)        {            while(l < r)            {                if(calc(l,r))  l++;                else { ans=r; break; }            }            r++;        }        printf("Case #%d: %d\n",Case++,ans);    }    return 0;}

 

HDU 5510---Bazinga（指针模拟）


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