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

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

Tehran 2002 Preliminary

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);
     }
 }