zoukankan      html  css  js  c++  java
  • PAT 甲级 1130 Infix Expression

    https://pintia.cn/problem-sets/994805342720868352/problems/994805347921805312

    Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20) which is the total number of nodes in the syntax tree. Then N lines follow, each gives the information of a node (the i-th line corresponds to the i-th node) in the format:

    data left_child right_child
    

    where data is a string of no more than 10 characters, left_child and right_child are the indices of this node's left and right children, respectively. The nodes are indexed from 1 to N. The NULL link is represented by −. The figures 1 and 2 correspond to the samples 1 and 2, respectively.

    infix1.JPGinfix2.JPG
    Figure 1 Figure 2

    Output Specification:

    For each case, print in a line the infix expression, with parentheses reflecting the precedences of the operators. Note that there must be no extra parentheses for the final expression, as is shown by the samples. There must be no space between any symbols.

    Sample Input 1:

    8
    * 8 7
    a -1 -1
    * 4 1
    + 2 5
    b -1 -1
    d -1 -1
    - -1 6
    c -1 -1
    

    Sample Output 1:

    (a+b)*(c*(-d))
    

    Sample Input 2:

    8
    2.35 -1 -1
    * 6 1
    - -1 4
    % 7 8
    + 2 3
    a -1 -1
    str -1 -1
    871 -1 -1
    

    Sample Output 2:

    (a*2.35)+(-(str%871))

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int N;
    int vis[30];
    
    struct Node{
        string val;
        int L, R;
    }node[30];
    
    string dfs(int st) {
        if(node[st].L == -1 && node[st].R == -1) return node[st].val;
        if(node[st].L == -1 && node[st].R != -1) return "(" + node[st].val + dfs(node[st].R) + ")";
        if(node[st].L != -1 && node[st].R != -1) return "(" + dfs(node[st].L) + node[st].val + dfs(node[st].R) + ")";
    }
    
    int main() {
        scanf("%d", &N);
        for(int i = 1; i <= N; i ++) {
            cin >> node[i].val >> node[i].L >> node[i].R;
            if(node[i].L != -1) vis[node[i].L] = 1;
            if(node[i].R != -1) vis[node[i].R] = 1;
        }
    
        int root = 1;
        while(vis[root] == 1) root ++;
        string ans = dfs(root);
        if(ans[0] == '(') ans = ans.substr(1, ans.size() - 2);
        cout << ans;
        return 0;
    }
    

      dfs 递归 最后去掉最外面的括号

    可能最近是自闭 girl 了 希望有好运气叭

  • 相关阅读:
    关于celery django djangocelery搭配报错问题及解决方法
    django 1048错误原因及解决思路
    CSS ::Selection
    Win7编程:在按钮中加入管理员权限运行盾牌图标转载
    VisualStudioVS2010统计代码行数
    在套用母版页的页面中应用input file上传图片
    Asp.Net Url 传值出现乱码的解决方法(包括js传值)
    JS验证码刷新无反应原因
    AspnetPager
    fckeditor2.6在IE9下的弹出窗口报错问题解决
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10417606.html
Copyright © 2011-2022 走看看