HDU-1238-Substrings(KMP, 暴力枚举)
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链接:
https://vjudge.net/problem/HDU-1238
题意:
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.
思路:
n个字符串的最长子串, 反着的也可以.
枚举第一个的子串, 对剩下的挨个匹配.
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 200;
const int MOD = 1e4+7;
string ori[MAXN], rev[MAXN];
int Next[MAXN];
int n;
void GetNext(string t)
int i = 0, k = -1;
int len = t.length();
Next[0] = -1;
while (i < len-1)
if (k == -1 || t[i] == t[k])
++i;
++k;
Next[i] = k;
else
k = Next[k];
bool Kmp(string s, string t)
int i = 0, j = 0;
int lens = s.length(), lent = t.length();
while (i < lens && j < lent)
if (j == -1 || s[i] == t[j])
++i;
++j;
else
j = Next[j];
return j >= lent;
int main()
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--)
cin >> n;
for (int i = 1;i <= n;i++)
cin >> ori[i];
rev[i].assign(ori[i].rbegin(), ori[i].rend());
int len = ori[1].length();
int ans = 0;
for (int i = 0;i < len;i++)
for (int j = i;j < len;j++)
string p(ori[1], i, j-i+1);
GetNext(p);
bool flag = true;
for (int k = 2;k <= n;k++)
if (!Kmp(ori[k], p) && !Kmp(rev[k], p))
flag = false;
break;
if (flag && (int)p.length() > ans)
ans = (int)p.length();
cout << ans << endl;
return 0;
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