HDU 1238字符串匹配

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11464 Accepted Submission(s): 5504

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

题解

1、题意就是找出n个字符串中相同的子串的长度(逆子串也可以)。

2、思路：
- 找到n个字符串中最小长度字符串
- 将最小长度字符串不断循环减短与另外的字符串比较
- 主要理解以下几个函数
- strlen:求字符串长度
- strncpy：strncpy()原型： char * strncpy(char *dest, const char *src, size_t n);
- strstr：strstr(str1,str2) 函数用于判断字符串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。
-
-

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

char s[101][101];
int m;
//反转
void rechange(char *str)
{
    int len;
    int j=0;
    char rs[101];
    len=strlen(str);
    for(int i=0;i<len;i++)
    {
        rs[j++]=str[len-i-1];
    }
    rs[j]='\0';
    strcpy(str,rs);
}
int GetStr(char *str)
{
    char p[101];//模拟子串
    char k[101];//模拟反转子串
    int len1,len2;
    int flag;
    len1=strlen(str);
    len2=strlen(str);
    while(len1>0)
    {
        for(int i=0;i<=len2-len1;i++)
        {
            /******/
            strncpy(p,str+i,len1);
            strncpy(k,str+i,len1);
            p[len1]='\0';
            k[len1]='\0';
            rechange(k);
            flag=1;
            for(int d=0;d<m;d++)
            {
                if(strstr(s[d],p)==NULL&&strstr(s[d],k)==NULL)
                {
                    flag=0;
                    break;
                }
            }
            if(flag==1)
            {
                return len1;
            }
        }
        len1--;
    }

}

int main()
{
    int n;
    char min_char[101];
    cin>>n;
    while(n--)
    {
        //int m;
        int min_length=100;
        cin>>m;
        for(int i=0;i<m;i++)
        {
            cin>>s[i];
            if(strlen(s[i])<min_length)
            {
                strcpy(min_char,s[i]);
                min_length=strlen(min_char);
            }
        }
        int sublen=GetStr(min_char);
        cout<<sublen<<endl;
    }
    return 0;
}