HDOJ 题目5510 Bazinga（瞎搞）

最新推荐文章于 2018-10-22 22:06:43 发布

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最新推荐文章于 2018-10-22 22:06:43 发布

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本文探讨了一种解决复杂字符串匹配问题的方法，通过分析样例输入和输出，提出了一个利用strstr函数进行字符串匹配的高效算法。该算法通过避免重复比较已知子串，显著提高了处理效率。详细阐述了算法原理、实现过程及性能优化，适用于类似比赛场景的快速字符串处理任务。

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Bazinga

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 331 Accepted Submission(s): 139

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

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

Recommend

wange2014 | We have carefully selected several similar problems for you: 5539 5538 5537 5536 5535

题目大意：就是让你找一个最大的i，是的最少存在一个0<j<i,字符串数sj不是si的子串

沈阳区域赛的题，太弱，并没有沈阳区域赛的资格，做重现的时候，感觉这个题不会很简单，就硬生生的写了200来行的在线ac自动机，，，死活超时。。原本想放弃了，看到讨论区里有说有简单地方法，可以用strstr过，于是就想了想，暴力肯定超时，它是从上往下有积累的很多串已经是一个串的子串了，和这个串比了，就没必要再和他的子串比了，，于是一顿瞎搞就过了。。第六，，

ac代码

1	Hacker_vision	156MS	2648K	1477B	G++	2015-10-31 18:00:50
2	_JM	171MS	2672K	1135B	G++	2015-10-31 17:53:30
3	戦い	218MS	2624K	1539B	G++	2015-10-31 21:54:18
4	马孟起	249MS	2644K	1040B	G++	2015-11-01 13:59:16
5	preserve	265MS	2672K	1026B	C++	2015-11-01 11:19:40
6	XY_	280MS	2716K	729B	C++	2015-11-01 20:41:08

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
char str[505][2002];
int next[2002];
int cc[505];
int n,m;
int main()
{
    int t,c=0;
    scanf("%d",&t);
    while(t--)
    {
        int num;
        scanf("%d",&num);
        int i,j,k;
        for(i=1;i<=num;i++)
            scanf("%s",str[i]);
        int ans=-1;
        int flag=0;
        for(i=num;i>1;i--)
            if(!strstr(str[i],str[i-1]))
            {
                ans=max(ans,i);
                for(j=i-1,k=i+1;k<=num;k++)
                    if(!strstr(str[k],str[j]))
                        ans=max(ans,k);
            }
        printf("Case #%d: %d\n",++c,ans);
    }
}