标签：des style http color os strong io 数据
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Expressions
Description Arithmetic expressions are usually written with the operators in between the two operands (which is called infix notation). For example, (x+y)*(z-w) is an arithmetic expression in infix notation. However, it is easier to write a program to evaluate an expression if the expression is written in postfix notation (also known as reverse Polish notation). In postfix notation, an operator is written behind its two operands, which may be expressions themselves. For example, x y + z w - * is a postfix notation of the arithmetic expression given above. Note that in this case parentheses are not required. To evaluate an expression written in postfix notation, an algorithm operating on a stack can be used. A stack is a data structure which supports two operations:
During the evaluation, we process the expression from left to right. If we encounter a number, we push it onto the stack. If we encounter an operator, we pop the first two numbers from the stack, apply the operator on them, and push the result back onto the stack. More specifically, the following pseudocode shows how to handle the case when we encounter an operator O: a := pop(); b := pop(); push(b O a); The result of the expression will be left as the only number on the stack. Now imagine that we use a queue instead of the stack. A queue also has a
Can you rewrite the given expression such that the result of the algorithm using the queue is the same as the result of the original expression evaluated using the algorithm with the stack? Input The first line of the input contains a number T (T ≤ 200). The following T lines each contain one expression in postfix notation. Arithmetic operators are represented by uppercase letters, numbers are represented by lowercase letters. You may assume that the length of each expression is less than 10000 characters. Output For each given expression, print the expression with the equivalent result when using the algorithm with the queue instead of the stack. To make the solution unique, you are not allowed to assume that the operators are associative or commutative. Sample Input 2 xyPzwIM abcABdefgCDEF Sample Output wzyxIPM gfCecbDdAaEBF Source |
题目大意：
解题思路：给你一个满二叉树的后序遍历，小写字母表示叶子节点，将这课树按照从下到上，从右到左顺序输出。
解题代码：先建立二叉树，用堆栈实现。
然后输出，用队列实现。
#include <iostream>
#include <string>
#include <cstdio>
#include <stack>
#include <queue>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=11000;
struct node{
int l,r;
char c;
}e[maxn];
int cnt;
char st[maxn];
void initial(){
int len=strlen(st);
for(int i=0;i<=len;i++){
e[i].l=e[i].r=-1;
}
cnt=0;
}
void solve(){
int len=strlen(st);
stack <int> v;
for(int i=0;i<len;i++){
if(st[i]>='a' && st[i]<='z'){
e[cnt].c=st[i];
v.push(cnt);
cnt++;
}else{
int r=v.top();
v.pop();
int l=v.top();
v.pop();
e[cnt].l=l;
e[cnt].r=r;
e[cnt].c=st[i];
v.push(cnt);
cnt++;
}
}
}
void output(){
string ans;
queue <int> q;
q.push(cnt-1);
while(!q.empty()){
int s=q.front();
q.pop();
ans.push_back(e[s].c);
if(e[s].l!=-1) q.push(e[s].l);
if(e[s].r!=-1) q.push(e[s].r);
}
reverse(ans.begin(),ans.end());
printf("%s\n",ans.c_str());
}
int main(){
int t;
scanf("%d",&t);
while(t-- >0){
scanf("%s",st);
initial();
solve();
output();
}
return 0;
}
POJ 3367 Expressions（数据结构-二叉树）,布布扣,bubuko.com
POJ 3367 Expressions（数据结构-二叉树）
标签：des style http color os strong io 数据
原文地址：http://blog.csdn.net/a1061747415/article/details/38299181
