HDU ACM 1238 Substrings

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6004    Accepted Submission(s): 2657

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

 

Recommend

Eddy

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define SIZE 102

using namespace std;

typedef struct strType
{
    char value[SIZE];
    int len;
}strType;

strType alpha[SIZE];
char pos[SIZE], neg[SIZE];

bool Traverse(int cur, int neglect, int n)
{
    if(cur == neglect) return Traverse(cur+1, neglect, n);
    if(cur == n) return true;
    
    if(strstr(alpha[cur].value, pos) != NULL || strstr(alpha[cur].value, neg) != NULL)
        return Traverse(cur+1, neglect, n);
    else return false;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, shortest, id;
        scanf("%d", &n);
        getchar();
        for(int i=0; i<n; ++i)
        {
            scanf("%s", alpha[i].value);
            alpha[i].len = strlen(alpha[i].value);
            if(!i) shortest = alpha[i].len, id = i;
            else if(shortest > alpha[i].len)
            {
                shortest = alpha[i].len;
                id = i;
            }
        }
        int res = 0;
        for(int i=alpha[id].len-1; i>=0; i--)
        {
            if(i+1 < res) break;
            for(int j=0; j<=i; j++)
            {
                if(i-j+1 < res) break;
                memcpy(pos, alpha[id].value+j, sizeof(char)*(i-j+1));
                neg[i-j+1] = pos[i-j+1] = '\0';
                for(int t=0; t<i-j+1; ++t) neg[i-j-t] = pos[t];
                if(Traverse(0, id, n))
                {
                    res = i-j+1;
                    if(res < i-j+1) res = i-j+1;
                }
            }
        }
        printf("%d\n", res);    
    }
    return 0;
}

【解题思路】strstr函数的应用，有几种不可能的情况需要先处理掉，所以一开始得找最短的字符串里的子字符串，且当前子字符串的长度至少要大于当前的符合所有字符串的要求  的子字符串