Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11914 Accepted Submission(s): 5708

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

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跟我上一篇博客异曲同工之妙

只要用strncpy 加两次kmp就可以判断

#include <cstdio>
#include <cstring>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <set>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
const int MAX_N = 65;
char str[15][MAX_N];
int Next[MAX_N],n,len;
int kmp(const char *T,const char *P){
    int t = -1,j = 0;
    Next[0] = -1;
    int lenT = strlen(T),lenP= strlen(P);
    while(j<lenP-1){
        if(t<0||P[t]==P[j]) {++t,++j,Next[j]=(P[j]==P[t]?Next[t]:t);}
        else t = Next[t];
    }
    int i = 0;
    j = 0;
    while(i<lenT&&j<lenP){
        if(j<0||T[i]==P[j]) i++,j++;
        else j = Next[j];
    }
    return j==lenP;
}
bool match(const char *Substr){
     char Substr_[124];
     int len = strlen(Substr);
     for(int i=0;i<len;i++) Substr_[i] = Substr[len-1-i];
     Substr_[len] = '\0';
     //dbg(Substr_);
     for(int i = 1;i<n;++i) if(!kmp(str[i],Substr)&&!kmp(str[i],Substr_)) return false;
     return true;
}
int main(){
    int t ;
    scanf("%d",&t);
    char ans[MAX_N],tmp[MAX_N];
    while(t--){
        memset(ans,0,sizeof(ans));
        scanf("%d\n",&n);
        for(int i=0;i<n;i++){
            scanf("%s",str[i]);
        }
            int flag = 0;
            int len_start = strlen(str[0]);
            for(len = len_start;len>0;--len){
                for(int pos = 0;pos<=len_start-len;pos++){
                    strncpy(tmp,str[0]+pos,len);
                    tmp[len] = '\0';
                    //dbg(tmp);
                    if(match(tmp) ) {
                        int lenT=strlen(tmp);
                        int lenA = strlen(ans);
                        if(lenT>lenA)
                            strcpy(ans,tmp);
                        flag = 1;
                    }
                }
                    if(flag) break;
        }
                //dbg(ans);
                int len_ans = strlen(ans);
                printf("%d\n",len_ans);

    }
    return 0;
}

HDOJ 1238 找子串匹配

Substrings

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