hdu1238 Substrings

最新推荐文章于 2020-03-19 19:22:56 发布

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最新推荐文章于 2020-03-19 19:22:56 发布

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分类专栏：字符串处理

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Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11563 Accepted Submission(s): 5560

Problem Description

You are given a number of case-sensitive(有大小写之分的) strings of alphabetic(字母的) characters, find the largest string X, such that either X, or itsinverse(相反的) 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

23ABCDBCDFFBRCD2roseorchid

Sample Output

22

Author

//题意：找出所有串的最长的公共连续子串
//思路：直接从最小的那串，枚举所有子串去寻找,直接枚举
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int main()  
{  
    int t,n,i,j,k,MIN,f,len,MAX;  
    char str[105][105],s1[105],s2[105];  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%d",&n);  
        MIN=1000;  
        for(i=0;i<n;i++)  
        {  
            scanf("%s",str[i]);  
            len=strlen(str[i]);  
            if(MIN>len)//找到最短串  
            {  
                MIN=len;  
                f=i;  
            }  
        }  
        len=strlen(str[f]);  
        int flag=1;  
        MAX=0;  
        for(i=0;i<len;i++)//作为最短串子串的头  
        {  
            for(j=i;j<len;j++)//子串的尾  
            {  
                for(k=i;k<=j;k++)//复制为两个串，顺序串s1，逆序串s2  
                {  
                    s1[k-i]=str[f][k];  
                    s2[j-k]=str[f][k];  
                }  
                s1[j-i+1]=s2[j-i+1]='\0'; //字符串的结束符 
                int l=strlen(s1);   
                for(k=0;k<n;k++)//将最短串的子串与其他所有字符串逐一对比  
                {   //函数str(str1,str2) 用于判断字符串str2是否是str1的子串。
					//如果是，则该函数返回str2在str1中首次出现的地址；否则，返回NULL。
                    if(!strstr(str[k],s1)&&!strstr(str[k],s2)) //顺序串和逆序串都不是该字符串的子串 
                    {  
                        flag=0;  
                        break;  
                    }  
                }  
                if(l>MAX&&flag)  
                MAX=l;  
                
                flag=1;//恢复标识，用于下次判断 
            }  
        }  
        printf("%d\n",MAX);  
    } 
    return 0;  
}