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ACM-字符串处理之Substrings——hdu1238

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6823    Accepted Submission(s): 3053

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


这道题目,可以用搜索来做。
我用的时string.h几个函数。

strcpy: 
char * strcpy(char * strDest,const char * strSrc);
就是将后面的数组赋给前面。如果前面数组长度不够大,运行崩溃。

strncpy:
char*strncpy(char*dest,char*src,size_tnum);
这个与strcpy差别在于,它可以控制长度,当然这两者都可以通过控制 第二个数组的头位置来控制赋值起点。

strlen:
extern unsigned int strlen(char *s);
这个就是返回s数组长度(不包括‘\0‘)

strstr:
extern char *strstr(const char *str1, const char *str2);
这个函数就是查找str2在str1中的位置,若不存在返回NULL。

仅这四个函数,即可解这道题。

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : Substrings                  *
*Source: hdu 1238                     *
* Hint  : 串的处理                   *
***************************************
**************************************/

#include <stdio.h>
#include <string.h>
char str[101][101];
int n;
int find_str(int len,int index)
{
    char s[101],pos[101],inv[101];
    bool flag;
    int i,j,l;
    //将字符串拷贝出来
    strcpy(s,str[index]);
    l=len;
    // 舍弃后面的
    while(l>0)
    {
        flag = 0;
        for(i = 0; i <= len-l; ++i)
        {
            flag=1;//flag初始化为1
            //正向字符串,将str中第i个位置,取l长度(舍弃前面的)
            strncpy(pos,s+i,l);
            //inv 为pos的逆序
            for(j=0; j<l; j++)
                inv[j]=pos[l-j-1];
            // 字符数组,最后\0,防止某些不必要的冲突
            pos[l]=inv[l] =‘\0‘;
            for(j=0; j<n; j++)
            {
                if(strstr(str[j],pos)==NULL&&strstr(str[j],inv)==NULL)
                {flag=0;break;}
            }
            if(flag)    break;
        }
        if(flag)    break;
        --l;
    }
    return l;
}

int main()
{

    int i,t,min_len,index;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        min_len = 101;
        for(i = 0; i < n; i++)
        {
            scanf("%s",str[i]);
            //找到最短的字符串
            if(strlen(str[i])<min_len)
            {
                min_len = strlen(str[i]);//记录最短字串的长度
                index = i;//记录最短字串的位置
            }
        }
        printf("%d\n",find_str(min_len,index));
    }
    return 0;
}


ACM-字符串处理之Substrings——hdu1238