HDU - 1238——Substrings （查找所有串中最长子串）

最新推荐文章于 2021-06-02 15:00:02 发布

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最新推荐文章于 2021-06-02 15:00:02 发布

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本文介绍了一种通过枚举和匹配方法寻找多个字符串中最长公共子串（包括其反转形式）的算法。该方法首先确定最短字符串，然后检查其所有可能的子串是否存在于其他字符串中。

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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.

InputThe 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.
OutputThere 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

题意:找出所有字符串中的公共串，倒过来也算。

思路:使用strstr（a,b）函数（可以匹配b中是否存在a），直接暴力枚举，找出最短的串，用最短的串中的子串暴力枚举每一个串。

#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
int t,n;
char a[150][110];
int find_ans(int len,int index)
{
    int flag;
    char S[110],pos[110],iny[110];
    strcpy(S,a[index]);
    int l=len;
    while(l>0)
    {
        for(int i=0; i<=len-l; i++)
        {
            flag=1;
            strncpy(pos,S+i,l);
            pos[l]='\0';
            for(int j=0; j<l; j++)
                iny[j]=pos[l-j-1];
            iny[l]='\0';
            for(int j=0;j<n;j++)
            {
                if(strstr(a[j],pos)==NULL&&strstr(a[j],iny)==NULL)
                {
                    flag=0;
                    break;
                }
            }
            if(flag)
                break;
        }
        if(flag)
            break;
        --l;
    }
    return l;

}
int main()
{
    int min_len,index;
    scanf("%d",&t);
    while(t--)
    {
        min_len=1000;
        scanf("%d",&n);
        for(int i=0; i<n; i++)
        {
            scanf("%s",a[i]);
            int len=strlen(a[i]);
            index=min_len>len?i:index;
            min_len=min_len>len?len:min_len;
        }

        printf("%d\n",find_ans(min_len,index));
    }
    return 0;
}