PAT 1130. Infix Expression (25) 折腾表达式，未解之谜

1130. Infix Expression (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

Sample Output 1:

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

Sample Input 2:

82.35 -1 -1* 6 1- -1 4% 7 8+ 2 3a -1 -1str -1 -1871 -1 -1

Sample Output 2:

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

所有的叶子节点都是数字。所有的非叶结点都是符号。一开始用叶子节点控制括号输出，各种繁琐和错误。改成用非叶结点控制括号输出后一下子就过了。

不过还有个疑问的地方就是为什么必须加上非叶节点的右子树是否为空的判断？？不加的话有个case过不了。非叶结点的右子树应该恒不空才对吧。不懂。

#include <iostream>#include<stdio.h>#include<queue>#include<algorithm>#include<vector>#include<string>#include<string.h>#include<set>#include<deque>#include<queue>using namespace std;struct{    string value;    int lc,rc;}node[60];int start;int mark[50]={0};void inorder(int x){     if(x==start)      {         if(node[x].lc!=-1)         inorder(node[x].lc);         cout<<node[x].value;         if(node[x].rc!=-1)         inorder(node[x].rc);         return ;      }     if(node[x].lc==-1&&node[x].rc==-1)     {           cout<<node[x].value;            return;     }     else     {         cout<<'(';         if(node[x].lc!=-1)         inorder(node[x].lc);         cout<<node[x].value;         if(node[x].rc!=-1)         inorder(node[x].rc);         cout<<')';     }}int main(){     int n;cin>>n;     for(int i=1;i<=n;i++)     {         string a;         int b,c;         cin>>a>>b>>c;         if(b!=-1) mark[b]=1;         if(c!=-1) mark[c]=1;         node[i].lc=b;         node[i].rc=c;         node[i].value=a;     }     start=1;     while(mark[start]==1) start++;     inorder(start);   return 0;}