Google Code Jam 2019 Round 1A 题解

A. Alien Rhyme

题意：

思路：将字符反向插入一颗Trie，然后自下而上的贪心即可，即先选后缀长的，再选后缀短的。

实现：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <iomanip>
  5 
  6 #include <vector>
  7 #include <cstring>
  8 #include <string>
  9 #include <queue>
 10 #include <deque>
 11 #include <stack>
 12 #include <map>
 13 #include <set>
 14 
 15 #include <utility>
 16 #include <list>
 17 
 18 #include <cmath>
 19 #include <algorithm>
 20 #include <cassert>
 21 #include <bitset>
 22 #include <complex>
 23 #include <climits>
 24 #include <functional>
 25 #include <unordered_set>
 26 #include <unordered_map>
 27 using namespace std;
 28 
 29 typedef long long ll;
 30 typedef pair<int, int> ii;
 31 typedef pair<ll, ll> l4;
 32 typedef pair<double, double> dd;
 33 #define mp make_pair
 34 #define pb push_back
 35 
 36 #define debug(x) cerr << #x << " = " << x << " "
 37 const int maxlen = 50+1;
 38 const int N = 1000+1;
 39 const int maxn = N * maxlen;
 40 const int A = 26;
 41 int g[maxn][A], cnt[maxn], tot;
 42 inline int newnode()
 43 {
 44     memset(g[tot], 0, sizeof(g[tot]));
 45     cnt[tot] = 0;
 46     return tot++;
 47 }
 48 void init()
 49 {
 50     tot = 0;
 51     newnode();
 52 }
 53 char s[maxlen];
 54 void insert()
 55 {
 56     int len = strlen(s);
 57     int cur = 0;
 58     for (int i = len-1; i >= 0; --i)
 59     {
 60         int c = s[i] - 'A';
 61         if (g[cur][c] == 0) g[cur][c] = newnode();
 62         cur = g[cur][c];
 63         ++cnt[cur];
 64     }
 65 }
 66 int solve(int cur)
 67 {
 68     if (cnt[cur] == 1) return 0;
 69     int ret = 0;
 70     for (int c = 0; c < A; ++c)
 71         if (g[cur][c])
 72         {
 73             //cerr << "from " << cur << " via " << char('A'+c) << " to " << g[cur][c] << endl;
 74             ret += solve(g[cur][c]);
 75         }
 76     int tmp = cnt[cur] - ret;
 77     if (tmp >= 2) ret += 2;
 78     return ret;
 79 }
 80 int main()
 81 {
 82 //    ios::sync_with_stdio(false);
 83 //    cin.tie(0);
 84 //    int T; cin >> T;
 85     int T; scanf("%d", &T);
 86     for (int kase = 1; kase <= T; ++kase)
 87     {
 88         int n; scanf("%d", &n);
 89         init();
 90         for (int i = 0; i < n; ++i)
 91         {
 92             scanf("%s", s);
 93             insert();
 94         }
 95         printf("Case #%d: %d\n", kase, solve(0));
 96     }
 97 }
 98 /*
 99  1
100  10 5 5
101  101010101000010100101
102  0 2 4 6 8 13 15  18 20
103  */

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