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 = 60000;
char s[maxn * 5], *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 = -1;
    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 != -1)
        {
            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 = 1; kase <= T;kase++)
    {
        dict.clear();
        cnt = 0;
        cin >> s;
        p = s;
        print(solve());  //返回根编号，递归输出
        cout << endl;
    }
    return 0;
}