hdu1238 Substrings

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1238

题目：

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10245    Accepted Submission(s): 4879

Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

 

 

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string. 

 

 

Output

There should be one line per test case containing the length of the largest string found.

 

 

Sample Input

2 3 ABCD BCDFF BRCD 2 rose orchid

 

 

Sample Output

2 2

 

 

Author

Asia 2002, Tehran (Iran), Preliminary

 

 

Recommend

Eddy

思路：暴力枚举+kmp

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int nt[K];
char sa[105][105],sb[105],sc[105];
void kmp_next(char *T,int *next)
{
    next[0]=0;
    for(int i=1,j=0,len=strlen(T);i<len;i++)
    {
        while(j&&T[i]!=T[j]) j=next[j-1];
        if(T[i]==T[j])  j++;
        next[i]=j;
    }
}
int kmp(char *S,char *T,int *next)
{
    int ans=0;
    int ls=strlen(S),lt=strlen(T);
    kmp_next(T,next);
    for(int i=0,j=0;i<ls;i++)
    {
        while(j&&S[i]!=T[j]) j=next[j-1];
        if(S[i]==T[j])  j++;
        if(j==lt)   ans++;
    }
    return ans;
}
int main(void)
{
    int t,n;cin>>t;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
            scanf("%s",sa[i]);
        int len=strlen(sa[1]),ans=0;
        for(int i=0;i<len;i++)
        {
            for(int j=i;j<len;j++)
            {
                int ff=1;
                sb[j-i]=sa[1][j],sb[j-i+1]='\0';
                if(j-i+1<=ans)continue;
                for(int k=j;k>=i;k--)
                    sc[j-k]=sa[1][k];
                sc[j-i+1]='\0';
                for(int k=2;k<=n&&ff;k++)
                if(!kmp(sa[k],sb,nt)&&!kmp(sa[k],sc,nt))
                    ff=0;
                if(ff)
                    ans=j-i+1;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}