UVa12219 Common Subexpression Elimination (表达式树)

Posted 2020-08-09

链接：http://vjudge.net/problem/UVA-12219

分析：用一个map把子树映射成编号1,2,...。这样一来，一棵子树就可以用三元组（s,left,right）表示（s表示根结点字符串，left，right表示左右子结点编号）。这样，每次判断一棵子树是否出现过只需要在map中查找，总时间复杂度为O（nlogn）。

 

 1 #include <cstdio>
 2 #include <string>
 3 #include <map>
 4 using namespace std;
 5 
 6 const int maxn = 5e4 + 5;
 7 
 8 int kase, cnt, done[maxn];
 9 char expr[maxn * 6], *p;
10 
11 struct Node {
12     string s;
13     int hash, left, right;
14     bool operator < (const Node& rhs) const {
15         if (hash != rhs.hash) return hash < rhs.hash;
16         if (left != rhs.left) return left < rhs.left;
17         return right < rhs.right;
18     }
19 } node[maxn];
20 
21 map<Node, int> dict;
22 
23 int parse() {
24     int id = ++cnt;
25     Node& u = node[id];
26     u.left = u.right = -1;
27     u.s = "";
28     u.hash = 0;
29     while (isalpha(*p)) {
30         u.hash = u.hash * 27 + *p - ‘a‘ + 1;
31         u.s.push_back(*p);
32         p++;
33     }
34     if (*p == ‘(‘) {
35         p++; u.left = parse(); p++; u.right = parse(); p++;
36     }
37     if (dict.count(u) != 0) {
38         cnt--;
39         return dict[u];
40     }
41     return dict[u] = id;
42 }
43 
44 void print(int u) {
45     if (done[u] == kase) printf("%d", u);
46     else {
47         done[u] = kase;
48         printf("%s", node[u].s.c_str());
49         if (node[u].left != -1) {
50             putchar(‘(‘);
51             print(node[u].left);
52             putchar(‘,‘);
53             print(node[u].right);
54             putchar(‘)‘);
55         }
56     }
57 }
58 
59 int main() {
60     int T;
61     scanf("%d", &T);
62     for (kase = 1; kase <= T; kase++) {
63         cnt = 0; dict.clear();
64         scanf("%s", &expr);
65         p = expr;
66         print(parse());
67         putchar(‘\n‘);
68     }
69     return 0;
70 }