UVa 12219 公共表达式消除

https://vjudge.net/problem/UVA-12219

题意：

用表达式树来表示一个表达式。

思路：

用map来记录出现过的子树。如(b,3,6)表示这棵子树的根为b，左子树为编号为3的子树，右子树为编号为6的子树。

 #include<iostream>
 #include<string>
 #include<cstring>
 #include<algorithm>
 #include<map>
 using namespace std;

 const int maxn = ;
 char s[maxn * ], *p;
 int cnt, kase;
 int vis[maxn];

 struct node
 {
     string s;   //因为运算数是1~4个小写字母，所以这里要用string
     int left, right;
     bool operator < (const node& rhs)  const     //有map，所以一定要重载
     {
         if (s != rhs.s)       return s < rhs.s;
         if (left != rhs.left)    return left < rhs.left;
         return right < rhs.right;
     }
 }tree[maxn];

 map<node, int>dict;

 int solve()
 {
     int id = ++cnt;  //从1开始编号
     node& t = tree[id];  //引用，比较方便
     t.s = "";
     t.left = t.right = -;
     while (*p >= 'a' && *p <= 'z')   //先把根输入进去
     {
         t.s.push_back(*p);
         p++;
     }

     //递归处理左右子树
     if (*p == '(')
     {
         p++;  //跳过'('
         t.left = solve(),  p++;     //跳过','
         t.right = solve(), p++;     //跳过')'
     }

     if (dict.count(t))   //如果已经出现过
     {
         cnt--;
         return dict[t];   //直接返回子树编号
     }
     return dict[t] = id;  //没出现过，编号并返回
 }

 void print(int u)
 {
     if (vis[u]==kase)     cout << u ;   //如果子树已经输出过，直接输出编号
     else
     {
         vis[u] = kase;
         cout << tree[u].s;
         //递归输出左右子树
         if (tree[u].left != -)
         {
             cout << "(";
             print(tree[u].left);
             cout << ",";
             print(tree[u].right);
             cout << ")";
         }
     }
 }

 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     int T;
     cin >> T;
     for (kase = ; kase <= T;kase++)
     {
         dict.clear();
         cnt = ;
         cin >> s;
         p = s;
         print(solve());  //返回根编号，递归输出
         cout << endl;
     }
     return ;
 }