HDU4287 字典树

最新推荐文章于 2020-08-26 17:18:11 发布

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最新推荐文章于 2020-08-26 17:18:11 发布

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原文链接：http://www.cnblogs.com/gj-Acit/p/4678583.html

本文介绍如何利用字典树与深度优先搜索解决特定问题，通过实例展示算法应用，包括初始化、插入字符串、搜索操作等步骤，旨在提供解决问题的新思路。

字典树建树，然后dfs搜索即可

  1 //#pragma comment(linker, "/STACK:1677721600")
  2 #include <map>
  3 #include <set>
  4 #include <stack>
  5 #include <queue>
  6 #include <cmath>
  7 #include <ctime>
  8 #include <vector>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <cstring>
 12 #include <cstdlib>
 13 #include <iostream>
 14 #include <algorithm>
 15 using namespace std;
 16 #define INF 0x3f3f3f3f
 17 #define inf (-((LL)1<<40))
 18 #define lson k<<1, L, (L + R)>>1
 19 #define rson k<<1|1,  ((L + R)>>1) + 1, R
 20 #define mem0(a) memset(a,0,sizeof(a))
 21 #define mem1(a) memset(a,-1,sizeof(a))
 22 #define mem(a, b) memset(a, b, sizeof(a))
 23 #define FIN freopen("in.txt", "r", stdin)
 24 #define FOUT freopen("out.txt", "w", stdout)
 25 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 26 #define dec(i, a, b) for(int i = a; i >= b; i --)
 27 
 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
 32 
 33 //typedef __int64 LL;
 34 typedef long long LL;
 35 const int MAXN = 100000 + 100;
 36 const int MAXM = 110000;
 37 const double eps = 1e-8;
 38 LL MOD = 1000000007;
 39 
 40 int t, n, m;
 41 char str[10], s1[5005][10];
 42 int tree[30010][27], s, ans;
 43 bool val[30010];//val[i] = 1表示是某个字符串的尾节点
 44 vector<int>e[12];
 45 
 46 void init() {
 47     int cnt = 5;
 48     rep (i, 'a', 'z') {
 49         cnt += 1 + (i == 's' || i == 'z' ? -1 : 0);
 50         e[cnt / 3].push_back(i - 'a');
 51     }
 52 }
 53 
 54 void insert_to_tree(char* str)
 55 {
 56     int u = 0;
 57     for(int i = 0; str[i]; i ++)
 58     {
 59         int c = str[i] - 'a';
 60         if(!tree[u][c])
 61         {
 62             tree[u][c] = ++s;
 63             val[s] = 0;
 64         }
 65         u = tree[u][c];
 66     }
 67     val[u] = 1;
 68 }
 69 
 70 void dfs(int u, int k, char *s, int n) {
 71     ans += val[u] && k == n;
 72     int p = s[k] - '0', sz = e[p].size();
 73     rep (i, 0, sz - 1) {
 74         int v = e[p][i];
 75         if(tree[u][v] && k < n)
 76             dfs(tree[u][v], k + 1, s, n);
 77     }
 78 }
 79 
 80 int main()
 81 {
 82 //    FIN;
 83     init();
 84     cin >> t;
 85     while(t--) {
 86         s = 0;
 87         mem0(tree); mem0(val);
 88         scanf("%d %d%*c", &n, &m);
 89         rep (i, 0, n - 1) scanf("%s", s1[i]);
 90         rep (i, 1, m) {
 91             scanf("%s", str);
 92             insert_to_tree(str);
 93         }
 94         rep (i, 0, n - 1) {
 95             ans = 0;
 96             dfs(0, 0, s1[i], strlen(s1[i]));
 97             cout << ans << endl;
 98         }
 99     }
100     return 0;
101 }