309 lines
8.6 KiB
C++
309 lines
8.6 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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/**
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从后缀表达式建立表达式树
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比如现在有以下一个后缀表达式：
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a b + c d e + * *
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根据这个后缀表达式建立二叉表达式树，算法如下：
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1. 依次读取表达式；
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2. 如果是操作数，则将该操作数压入栈中；
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3. 如果是操作符，则弹出栈中的两个操作数，第一个弹出的操作数作为右孩子，第二个弹出的操作数作为左孩子；然后再将该操作符压入栈中。
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这样下去，就可以建立一颗完整的表达式树。
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*/
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// https://www.cnblogs.com/Running-Time/p/4922670.html
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typedef long long LL;
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const int N = 1e5 + 10;
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const int INF = 0x3f3f3f3f;
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const int MOD = 1e9 + 7;
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/*
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表达式求值，逆波兰式(后缀表达式)算法
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输入(可以有空格，支持小数，实现'+-/*%')： ((1+2)*5+1)/4=
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注意：取模一定是要整型，实现版本数字全是double，强制类型转换可能倒置错误
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转换为后缀表达式： 得到：1 2 + 5 * 1 + 4 / =
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计算后缀表达式：得到：4.00
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*/
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bool is_digit(char ch) {
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return '0' <= ch && ch <= '9';
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}
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struct Exp {
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stack<char> op;
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stack<double> num;
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int prior(char ch) { //运算符的优先级
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switch (ch) {
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case '+':
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case '-':
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return 1;
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case '*':
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case '%':
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case '/':
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return 2;
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default:
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return 0;
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}
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}
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//中缀表达式转变前缀表达式
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string get_prefix(string s) {
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while (!op.empty()) op.pop();
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op.push('#');
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string ret = "";
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int len = s.length(), i = len - 1;
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while (i >= 0) {
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if (s[i] == ' ' || s[i] == '=') {
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i--;
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continue;
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} else if (s[i] == ')') {
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op.push(s[i--]);
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} else if (s[i] == '(') {
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while (op.top() != '#' && op.top() != ')') {
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ret += op.top();
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ret += ' ';
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op.pop();
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}
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op.pop();
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i--;
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} else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/' || s[i] == '%') {
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while (prior(op.top()) > prior(s[i])) {
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ret += op.top();
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ret += ' ';
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op.pop();
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}
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op.push(s[i--]);
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} else {
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while (is_digit(s[i]) || s[i] == '.') {
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ret += s[i--];
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}
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ret += ' ';
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}
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}
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while (op.top() != '#') {
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ret += op.top();
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ret += ' ';
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op.pop();
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}
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reverse(ret.begin(), ret.end());
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return ret;
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}
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//计算
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double cal(double a, double b, char ch) {
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if (ch == '+') return a + b;
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if (ch == '-') return a - b;
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if (ch == '*') return a * b;
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if (ch == '%') return (int) ((int) a % (int) b);
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if (ch == '/') {
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if (b != 0) return a / b;
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return 0;
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}
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return 0;
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}
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//中缀表达式转变后缀表达式
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string get_postfix(string s) {
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while (!op.empty()) op.pop();
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op.push('#');
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string ret = "";
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int len = s.length(), i = 0;
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while (i < len) {
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if (s[i] == ' ' || s[i] == '=') {
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i++;
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continue;
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} else if (s[i] == '(') {
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op.push(s[i++]);
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} else if (s[i] == ')') {
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while (op.top() != '(') {
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ret += op.top();
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ret += ' ';
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op.pop();
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}
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op.pop();
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i++;
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} else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/' || s[i] == '%') {
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while (prior(op.top()) >= prior(s[i])) {
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ret += op.top();
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ret += ' ';
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op.pop();
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}
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op.push(s[i++]);
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} else {
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while (is_digit(s[i]) || s[i] == '.') {
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ret += s[i++];
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}
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ret += ' ';
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}
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}
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while (op.top() != '#') {
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ret += op.top();
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ret += ' ';
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op.pop();
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}
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ret += '=';
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return ret;
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}
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//计算后缀表达式
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double solve(string str) {
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string s = get_postfix(str);
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while (!num.empty()) num.pop();
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int len = s.length(), i = 0;
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while (i < len) {
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if (s[i] == ' ' || s[i] == '=') {
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i++;
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continue;
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} else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/' || s[i] == '%') {
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double a = num.top();
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num.pop();
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double b = num.top();
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num.pop();
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num.push(cal(b, a, s[i]));
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i++;
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} else {
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double x = 0;
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while (is_digit(s[i])) {
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x = x * 10 + s[i] - '0';
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i++;
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}
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if (s[i] == '.') {
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double k = 10.0, y = 0;
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i++;
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while (is_digit(s[i])) {
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y += ((s[i] - '0') / k);
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i++;
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k *= 10;
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}
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x += y;
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}
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num.push(x);
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}
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}
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return num.top();
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}
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} E;
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typedef struct BT {
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char op;
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double data;
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BT *lch, *rch;
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} node, *btree;
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void print(btree p) {
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if (!(p->lch) && !(p->rch)) {
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cout << fixed << setprecision(1) << p->data << " ";
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} else cout << p->op << " ";
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}
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int i;
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double get_num(string s) {
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double x = 0;
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while (is_digit(s[i])) {
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x = x * 10 + s[i] - '0';
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i++;
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}
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if (s[i] == '.') {
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double k = 10.0, y = 0;
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i++;
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while (is_digit(s[i])) {
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y += ((s[i] - '0') / k);
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i++;
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k *= 10;
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}
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x += y;
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}
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return x;
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}
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void create(btree &T, string s) {
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T = new node;
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while (i < s.length() && (s[i] == ' ' || s[i] == '.')) i++;
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if (i >= s.length()) {
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T->lch = T->rch = NULL;
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return;
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}
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if (is_digit(s[i])) {
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T->data = get_num(s);
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T->lch = T->rch = NULL;
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i++;
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return;
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} else {
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T->op = s[i];
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i += 2;
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create(T->lch, s);
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create(T->rch, s);
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}
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}
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void pre_order(btree T) { //前序遍历
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if (T != NULL) {
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print(T);
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pre_order(T->lch);
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pre_order(T->rch);
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}
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}
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void in_order(btree T) { //中序遍历
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if (T != NULL) {
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in_order(T->lch);
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print(T);
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in_order(T->rch);
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}
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}
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void post_order(btree T) { //后序遍历
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if (T != NULL) {
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post_order(T->lch);
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post_order(T->rch);
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print(T);
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}
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}
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double cal_tree(btree &T) {
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if (T != NULL) {
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if (!(T->lch) && !(T->rch)) {
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return T->data;
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} else return E.cal(cal_tree(T->lch), cal_tree(T->rch), T->op);
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}
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}
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int main() {
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int T;
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cin >> T;
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string str;
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getline(cin, str);
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while (T--) {
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getline(cin, str);
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string pre = E.get_prefix(str), post = E.get_postfix(str);
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cout << "前缀表达式：" << pre << endl;
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cout << "后缀表达式：" << post << endl;
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btree tree;
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i = 0; //全局变量记录string下标，递归建立表达式树
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create(tree, pre);
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cout << "前序遍历：";
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pre_order(tree);
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cout << endl;
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cout << "中序遍历：";
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in_order(tree);
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cout << endl;
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cout << "后序遍历：";
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post_order(tree);
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cout << endl;
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cout << "后缀表达式计算结果：" << fixed << setprecision(6) << E.solve(str) << endl;
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cout << "表达式树结果：" << fixed << setprecision(6) << cal_tree(tree) << endl << endl;
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}
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return 0;
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}
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/*
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测试样例：
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1000
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1 + 2
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1 + 2 - 3 / 4 * 5
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1 + 2 * (3 - 4) - 5 / 6
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1+2*3-4/(5-6*7+8)/9
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1-2+3*4%(5+6-7/8)+9
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1+(2-3)*4*(5-6+7/8)/9
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11+(22-33)*44*(55-6+7/8)/9
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19+(28-37)*46*(55-64+73/82)/91
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1.2+(2.421-3.3123)*4.0*(5.42342-6+7.2*8.13)/9.1321
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*/ |