POJ1226 Substrings ——后缀数组 or 暴力+strstr()函数 最长公共子串

时间：2025-05-07 14:33:38

题目链接：https://vjudge.net/problem/POJ-1226

Substrings
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 15122   Accepted: 5309

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 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

Source

题意：

给出n个字符串，问是否存在一个公共子串存在于每个字符串或其逆串中，若存在，输出最长的长度。

题解：

1.将所有字符串已经其逆串拼接在一起，相邻两个之间用各异的分隔符隔开。

2.求出新串的后缀数组，然后二分公共子串的长度mid：mid将新串的后缀分成若干组，每一组对应着一个公共子串，并且长度>=mid。如果这组出现了所有字符串，那么说明当前mid合法，否则不合法。最终求得答案。

3.由于数据较弱，还可以直接暴力+kmp/strstr()函数。

后缀数组：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <cmath>

 #include <queue>

 #include <stack>

 #include <map>

 #include <string>

 #include <set>

 using namespace std;

 typedef long long LL;

 const int INF = 2e9;

 const LL LNF = 9e18;

 const int MOD = 1e9+;

 const int MAXN = 1e5+;

 int id[MAXN];

 int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];

 int t1[MAXN], t2[MAXN], c[MAXN];

 bool cmp(int *r, int a, int b, int l)

 {

     return r[a]==r[b] && r[a+l]==r[b+l];

 }

 void DA(int str[], int sa[], int Rank[], int height[], int n, int m)

 {

     n++;

     int i, j, p, *x = t1, *y = t2;

     for(i = ; i<m; i++) c[i] = ;

     for(i = ; i<n; i++) c[x[i] = str[i]]++;

     for(i = ; i<m; i++) c[i] += c[i-];

     for(i = n-; i>=; i--) sa[--c[x[i]]] = i;

     for(j = ; j<=n; j <<= )

     {

         p = ;

         for(i = n-j; i<n; i++) y[p++] = i;

         for(i = ; i<n; i++) if(sa[i]>=j) y[p++] = sa[i]-j;

         for(i = ; i<m; i++) c[i] = ;

         for(i = ; i<n; i++) c[x[y[i]]]++;

         for(i = ; i<m; i++) c[i] += c[i-];

         for(i = n-; i>=; i--) sa[--c[x[y[i]]]] = y[i];

         swap(x, y);

         p = ; x[sa[]] = ;

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

             x[sa[i]] = cmp(y, sa[i-], sa[i], j)?p-:p++;

         if(p>=n) break;

         m = p;

     }

     int k = ;

     n--;

     for(i = ; i<=n; i++) Rank[sa[i]] = i;

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

     {

         if(k) k--;

         j = sa[Rank[i]-];

         while(str[i+k]==str[j+k]) k++;

         height[Rank[i]] = k;

     }

 }

 bool vis[];

 bool test(int n, int len, int k)

 {

     int cnt = ;

     memset(vis, false, sizeof(vis));

     for(int i = ; i<=len; i++)

     {

         if(height[i]<k)

         {

             cnt = ;

             memset(vis, false, sizeof(vis));

         }

         else

         {

             if(!vis[id[sa[i-]]]) vis[id[sa[i-]]] = true, cnt++;

             if(!vis[id[sa[i]]]) vis[id[sa[i]]] = true, cnt++;

             if(cnt==n) return true;

         }

     }

     return false;

 }

 char str[MAXN];

 int main()

 {

     int T, n;

     scanf("%d", &T);

     while(T--)

     {

         scanf("%d", &n);

         int len = ;

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

         {

             scanf("%s", str);

             int LEN = strlen(str);

             for(int j = ; j<LEN; j++)

             {

                 r[len] = str[j];

                 id[len++] = i;

             }

             r[len] = +i; //分隔符

             id[len++] = i;

         }

         for(int i = ; i<len; i++)  //逆串

         {

             r[*len--i] = r[i];

             if(r[*len--i]>=) r[*len--i] += ;  //分隔符应各异

             id[*len--i] = id[i];

         }

         len *= ;

         r[len] = ;

         DA(r,sa,Rank,height,len,);

         int l = , r = ;

         while(l<=r)

         {

             int mid = (l+r)>>;

             if(test(n,len,mid))

                 l = mid + ;

             else

                 r = mid - ;

         }

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

     }

 }

暴力 + strstr()函数：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cstdlib>

 #include <string>

 #include <vector>

 #include <map>

 #include <set>

 #include <queue>

 #include <sstream>

 #include <algorithm>

 using namespace std;

 typedef long long LL;

 const double eps = 1e-;

 const int INF = 2e9;

 const LL LNF = 9e18;

 const int MOD = 1e9+;

 const int MAXN = +;

 char s[MAXN][MAXN], t1[MAXN], t2[MAXN];

 int main()

 {

     int T, n;

     scanf("%d", &T);

     while(T--)

     {

         scanf("%d", &n);

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

             scanf("%s", s[i]);

         int Len = strlen(s[]), ans = ;

         for(int len = Len; len>=; len--)

         {

             for(int st = ; st<=Len-len; st++)

             {

                 int en = st+len-, cnt = ;

                 for(int i = st; i<=en; i++)

                 {

                     t1[cnt] = s[][i];

                     t2[cnt++] = s[][i];

                 }

                 t1[cnt] = t2[cnt] = ;

                 reverse(t2,t2+cnt);

                 bool flag = true;

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

                     flag = flag&&(strstr(s[i], t1) || strstr(s[i], t2) );

                 if(flag)

                 {

                     ans = len;

                     break;

                 }

             }

             if(ans==len) break;

         }

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

     }

 }

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