zoukankan      html  css  js  c++  java
  • A1130. Infix Expression

    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 -1. The figures 1 and 2 correspond to the samples 1 and 2, respectively.

    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))

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<string>
     5 using namespace std;
     6 typedef struct{
     7     int lchild, rchild;
     8     string data;
     9 }node;
    10 node tree[21];
    11 int N, notRoot[21] = {0};
    12 void inOrder(int root, int first){
    13     if(root == -1)
    14         return;
    15     string ss = tree[root].data;
    16     if(tree[root].rchild != -1 && first == 0)
    17         cout << "(";
    18     inOrder(tree[root].lchild, 0);
    19     cout << ss;
    20     inOrder(tree[root].rchild, 0);
    21     if(tree[root].rchild != -1 && first == 0)
    22         cout << ")";
    23 }
    24 int main(){
    25     scanf("%d", &N);
    26     for(int i = 1; i <= N; i++){
    27         int cl, cr;
    28         string ss;
    29         cin >> ss >> cl >> cr;
    30         tree[i].lchild = cl;
    31         tree[i].rchild = cr;
    32         tree[i].data = ss;
    33         if(cl != -1)
    34             notRoot[cl] = 1;
    35         if(cr != -1)
    36             notRoot[cr] = 1;
    37     }
    38     int root;
    39     for(root = 1; root <= N && notRoot[root] == 1; root++);
    40     inOrder(root, 1);
    41     cin >> N;
    42     return 0;
    43 }
    View Code

    总结:

    1、中序遍历即可,当该节点不是叶节点时,则需要先输出" ( ", 再递归访问左子树,右括号同理。起初我是用判断 + - * %来判断是否是非叶节点,结果最后一个测试点过不去。后来改成判断子树是否为空就全过了。这里可能是由于运算符号不止有+-*%,可能还有未知的运算符号。

  • 相关阅读:
    刷新界面
    分页加载数据(每次10条内容)的简单计算
    Intent传输包含对象的List集合
    android定时更新文件
    java中byte数据转换为c#的byte数据
    java zip文件的解压缩(支持中文文件名)
    Redis PHP扩展安装步骤
    CentOS6.5 开机启动自动运行redis服务
    centos7.2挂载硬盘攻略
    探究:Adobe Premiere Pro CC 2018 导入SRT字幕显示不全问题
  • 原文地址:https://www.cnblogs.com/zhuqiwei-blog/p/8586916.html
Copyright © 2011-2022 走看看