Poj 3294 Life Forms (后缀数组 + 二分 + Hash)

题目链接:

　　Poj 3294 Life Forms

题目描述：

　　有n个文本串，问在一半以上的文本串出现过的最长连续子串？

解题思路：

　　可以把文本串用没有出现过的不同字符连起来，然后求新文本串的height。然后二分答案串的长度K，根据K把新文本串的后缀串分块，统计每块中的原文本串出现的次数，大于原文本串数目的一半就作为答案记录下来，对于输出字典序，height就是排好序的后缀数组，只要按照顺序输出即可。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 const int maxn = ;

 int sa[maxn], rank[maxn], height[maxn], vis[], res[maxn];
 int t1[maxn], t2[maxn], r[maxn], flag[maxn], c[maxn];

 bool cmp (int *str, int a, int b, int k)
 {
     return str[a]==str[b] && str[a+k]==str[b+k];
 }

 void da (int *str, int n, int m)
 {
     n ++;
     int *x = t1, *y = t2, i, j;

     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*=)
     {
         int 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 (int i=; i<n; i++)//i是rank
             x[sa[i]] = cmp(y, sa[i-], sa[i], j)?p-:p++;
         if (p >= n)
             break;
         m = p;
     }

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

     int k = ;
     n --;
     for (int i=; i<n; i++)
     {
         if (k)  k --;
         int j = sa[rank[i] - ];
         while (str[i+k] == str[j+k])    k++;
         height[rank[i]] = k;
     }
 }

 bool Bin_sreach (int x, int k, int n)
 {
     int ans, num;
     ans = num = ;
     memset (vis, , sizeof(vis));

     for (int i=; i<=k; i++)
     {
         if (height[i] >= x)
         {
             ans += vis[flag[sa[i-]]]?:;
             vis[flag[sa[i-]]] = ;

             ans += vis[flag[sa[i]]]?:;
             vis[flag[sa[i]]] = ;
         }
         else
         {
             if (ans* > n)
                 res[++ num] = sa[i-];

             ans = ;
             memset (vis, , sizeof(vis));
         }
     }
     if (ans* > n)
         res[++ num] = sa[k-];

     if (num)
     {
         res[] = num;
         return true;
     }
     return false;
 }

 int main ()
 {
     int n, l = ;
     char str[];
     while (scanf ("%d", &n), n)
     {
         if (l ++)
             printf ("\n");

         int k = ;
         for (int i=; i<n; i++)
         {
             scanf ("%s", str);
             for (int j=; str[j]; j++)
             {
                 r[k] = str[j];
                 flag[k++] = i;//记录k字母所在的字符串
             }
             r[k] =  + i;
             flag[k++] = -;
         }

         r[k] = ;
         da (r, k, );

         int low = , high = k, mid, ans = ;
         while (low <= high)
         {//二分枚举
             mid = (low + high) / ;
             if (Bin_sreach(mid, k, n))
             {
                 ans = mid;
                 low = mid + ;
             }
             else
                 high = mid - ;
         }

         if (low == )
         {
             printf ("?\n");
             continue;
         }

         for (int i=; i<=res[]; i++)
             {
                 for (int j=res[i]; j<res[i]+ans; j ++)
                     printf ("%c", r[j]);
                 printf ("\n");
             }
     }
     return ;
 }