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Source : ACM ICPC Tehran 2002 Preliminery ContestTime limit : 1 secMemory limit : 32 M

Submitted : 568, Accepted : 191

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 

23ABCDBCDFFBRCD2roseorchid

Sample Output 

22 

题意：给一系列字符串，找出一个字符串X使得任意给的字符串里面都能找到X或者是反过来的X，然后要输出最大的X的长度，这里有点坑 ，是字符的值得和最大的长度，而不是长度最长的。

思路： 暴搜吧  反正也是0ms过

代码:

#include<iostream>

#include<cstdio>

#include<string.h>

#include<algorithm>

#include<string>

#include<deque>

#include<queue>

#include<math.h>

#include<vector>

using namespace std;

#define MAX 100000+10

#define MOD 99997

const int inf = 0xfffffff;

char input[110][110];

int len[110];

int n;

bool isOk(char  s[] , int s_len)

{

int move = 1;

bool flag;

for (int i = 2 ; i <= n ; ++i)

{

flag = false;

for (int j = 1 ; j <= len[i]-s_len+1 ; ++j) if (input[i][j]==s[1])

{

int k = 1;

while (k<=s_len && s[k]==input[i][j+k-1]) ++k;

if (k==s_len+1)

{

flag = true;

break;

}

}

if (flag) continue;

for (int j = 1 ; j <= len[i]-s_len+1 ; ++j) if (input[i][j]==s[s_len])

{

int k = 1;

while (k<=s_len && input[i][j+k-1]==s[s_len-k+1]) ++k;

if (k>s_len)

{

flag = true;

break;

}

}

if (!flag) return false;

}

return true;

}

int main()

{

int T;

cin>>T;

while (T--)

{

scanf("%d",&n);

int k = 1;

for (int i = 1 ; i <= n ; ++i)

{

scanf("%s",input[i]+1);

len[i] = strlen(input[i]+1);

k = !k;

}

bool  flag = false;

k = !k;

char X[110];

int X_len = 0;

int largest = -inf;

int ans = 0;

for (int i = 1 ; i <= len[1] ; ++i)

{

X_len = 0;

int j = i;

int sum = 0;

while (j <= len[1]) 

{

X[++X_len] = input[1][j];

sum += input[1][j];

if (sum>largest && isOk(X,X_len))

{

largest = sum;

ans = X_len;

}

++j;

}

}

printf("%d\n",ans);

}

}