hdoj 5510 Bazinga 【不要想太多。。。 strstr】

最新推荐文章于 2019-05-21 17:00:15 发布

原创最新推荐文章于 2019-05-21 17:00:15 发布 · 768 阅读

0 ·

CC 4.0 BY-SA版权

水题专栏收录该内容

127 篇文章

订阅专栏

本文探讨了在给定一组字符串的情况下，寻找特定条件下的最大编号。通过详细解析输入输出样例，阐述了一种有效算法解决子串匹配问题，特别关注于识别不存在于特定位置之前的字符串作为子串的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Bazinga

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 72 Accepted Submission(s): 28

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：给你n个串，让你找到最大的i(字符串编号从1开始)使得至少存在一个j(1<=j<i)满足串j不是串i的子串。

不要想太多，弱渣ac自动机TLE到死 o(╯□╰)o

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#define lson o<<1|1, l, mid
#define rson o<<1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define INF 0x3f3f3f3f
#define eps 1e-8
#define debug printf("1\n")
#define MAXN 100010
#define MAXM 20000000
#define LL long long
#define CLR(a, b) memset(a, (b), sizeof(a))
#define W(a) while(a--)
#define Ri(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define Rl(a) scanf("%lld", &a)
#define Pl(a) printf("%lld\n", (a))
#define Rs(a) scanf("%s", a)
#define Ps(a) printf("%s\n", (a))
#define MOD 1000000007
#define LL long long
using namespace std;
char str[510][2010];
int main()
{
    int t, kcase = 1;
    Ri(t);
    W(t)
    {
        int n; Ri(n);
        for(int i = 0; i < n; i++)
            Rs(str[i]);
        int ans = -2;
        for(int i = n-1; i > 0; i--)
        {
            if(!strstr(str[i], str[i-1]))
            {
                ans = max(ans, i);
                for(int j = i+1; j < n; j++)
                    if(!strstr(str[j], str[i-1]))
                        ans = max(ans, j);
            }
        }
        printf("Case #%d: %d\n", kcase++, ans+1);
    }
    return 0;
}