BZOJ - 3277 - 串

题意

现在给定你n个字符串，询问每个字符串有多少子串（不包括空串）是所有n个字符串中至少k个字符串的子串（注意包括本身）。

题解

首先是广义后缀自动机，就是每次将$last$归为$Root$，从而将几个后缀自动机拼在一起处理

那么现在需要知道每个字串在$n$个母串中的出现次数，所谓字串，就是所有前缀的所有后缀，所以可以顺着前缀走，那么通过$Parent$树找后缀，一直往上跳，那么需要加一的所有后缀就是所属母串并非当前母串的那几个

此时再整理出每个所属母串个数$Size >= K$的初始贡献，即$Len[i] - Len[Father[i]]$，反之赋$0$

又子串为前缀的后缀，那么一个节点的贡献即为它祖先至它本身的贡献前缀和，即它所有后缀能够构成的子串数

最后再遍历一遍前缀统计答案即可

代码

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <string>
  5 #include <vector>
  6 
  7 using namespace std;
  8  
  9 typedef long long LL;
 10  
 11 const int MAXN = 6e05 + 10;
 12  
 13 int N, K;
 14 string Str[MAXN >> 2];
 15 
 16 const int Root = 1;
 17 int Tree[MAXN][30]= {0};
 18 int Father[MAXN]= {0};
 19 int Len[MAXN]= {0}, Size[MAXN]= {0};
 20 int Belong[MAXN]= {0};
 21 int last = Root, nodes = 1;
 22 void Append (int c, int bel) {
 23     int fa = last, p = ++ nodes;
 24     last = p;
 25     Len[p] = Len[fa] + 1;
 26     while (fa && ! Tree[fa][c])
 27         Tree[fa][c] = p, fa = Father[fa];
 28     if (! fa)
 29         Father[p] = 1;
 30     else {
 31         int x = Tree[fa][c];
 32         if (Len[x] == Len[fa] + 1)
 33             Father[p] = x;
 34         else {
 35             int np = ++ nodes;
 36             Len[np] = Len[fa] + 1, Father[np] = Father[x];
 37             Father[p] = Father[x] = np;
 38             memcpy (Tree[np], Tree[x], sizeof (Tree[x]));
 39             while (fa && Tree[fa][c] == x)
 40                 Tree[fa][c] = np, fa = Father[fa];
 41         }
 42     }
 43 }
 44 int Topo[MAXN];
 45 int W[MAXN]= {0};
 46 int buck[MAXN]= {0};
 47 void Merge () {
 48     for (int i = 1; i <= N; i ++) {
 49         int n = Str[i].size();
 50         int p = Root;
 51         for (int j = 0; j < n; j ++) {
 52             p = Tree[p][Str[i][j] - 'a'];
 53             int fa = p;
 54             while (fa && Belong[fa] != i) {
 55                 Size[fa] ++;
 56                 Belong[fa] = i;
 57                 fa = Father[fa];
 58             }
 59         }
 60     }
 61     for (int i = 1; i <= nodes; i ++)
 62         buck[Len[i]] ++;
 63     for (int i = 1; i <= nodes; i ++)
 64         buck[i] += buck[i - 1];
 65     for (int i = nodes; i >= 1; i --)
 66         Topo[buck[Len[i]] --] = i;
 67     for (int i = 1; i <= nodes; i ++)
 68         if (Size[i] >= K)
 69             W[i] = Len[i] - Len[Father[i]];
 70     for (int i = 1; i <= nodes; i ++)
 71         W[Topo[i]] += W[Father[Topo[i]]];
 72 }
 73  
 74 LL Answer[MAXN >> 2]= {0};
 75  
 76 int main () {
 77     scanf ("%d%d", & N, & K);
 78     for (int i = 1; i <= N; i ++) {
 79         cin >> Str[i];
 80         int n = Str[i].size();
 81         last = Root;
 82         for (int j = 0; j < n; j ++)
 83             Append (Str[i][j] - 'a', i);
 84     }
 85     Merge ();
 86     for (int i = 1; i <= N; i ++) {
 87         int n = Str[i].size();
 88         int p = Root;
 89         for (int j = 0; j < n; j ++)
 90             p = Tree[p][Str[i][j] - 'a'], Answer[i] += W[p];
 91     }
 92     for (int i = 1; i <= N; i ++) {
 93         if (i > 1)
 94             putchar (' ');
 95         printf ("%lld", Answer[i]);
 96     }
 97     puts ("");
 98  
 99     return 0;
100 }
101  
102 /*
103 3 1
104 abc
105 a
106 ab
107 */
108 
109 /*
110 2 2
111 aa
112 aaa
113 */
View Code

 

posted @ 2018-12-07 20:51  Colythme  阅读(118)  评论(0)  编辑  收藏  举报