算法竞赛入门经典（第二版-刘汝佳） Uva442 经验总结

题目链接

Sample Input
9
A 50 10
B 10 20
C 20 5
D 30 35
E 35 15
F 15 5
G 5 10
H 10 20
I 20 25
A
B
C
(AA)
(AB)
(AC)
(A(BC))
((AB)C)
(((((DE)F)G)H)I)
(D(E(F(G(HI)))))
((D(EF))((GH)I))

Sample Output
0
0
0
error
10000
error
3500
15000
40500
47500
15125

一开始的思路是用链表存放每个矩阵的行、列、字母，然后把字母入栈，根据字母来找对应的矩阵，程序写出来后虽然能过样例，但是却无法AC，而且代码又长，时间并不是很宽裕就放弃原先的思路去学习刘汝佳老师的代码

自己写的代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
using namespace std;
typedef struct multi* mt;
struct multi
{
	int m;
	int n;
	char c;
	mt next;
	multi() : next(NULL) {}
};
stack<char> pt;
char order;
int sum = 0, flag = 0, n, nl;
mt head = new struct multi(), p1 = NULL;

mt find(char c) {//根据字母来查找对应的矩阵
	p1 = head;
	while (p1 != NULL) {
		if (p1->c == c) {
			return p1;
		}
		p1 = p1->next;
	}
}

void Insert(mt temp) { //新乘出来的矩阵插入链表末尾
	p1 = head;
	while (p1->next != NULL) {
		p1 = p1->next;
	}
	p1->next = temp;
}

void Clear() {  //有关数据全部初始化
	sum = 0;
	order = 'a'; // 给新的矩阵赋予字母编号，便于查找
	while (!pt.empty()) {
		pt.pop();
	}
	int j = 0;
	p1 = head->next;
	while (j < nl - 1) {
		p1 = p1->next;
		j++;
	}
	if (p1 != NULL)
		p1->next = NULL;
}

/*void test() {
	cout << "---------------" << endl;
	p1 = head->next;
	while (p1 != NULL) {
		cout << p1->c << " " << p1->m << " " << p1->n << endl;
		p1 = p1->next;
	}
	cout << "----------------" << endl;
}*/

int main() {
	//freopen("D://in.txt", "r", stdin);
	ios::sync_with_stdio(0); 
	p1 = head;
	cin >> n;
	nl = n;
	while (n--) {
		mt p2 = new struct multi();
		cin >> p2->c >> p2->m >> p2->n;
		p1->next = p2;
		p1 = p2;
	}
	getchar();
	string cmd;
	while (getline(cin, cmd)) {
		if (cmd.length() == 1) {
			cout << "0" << endl;
			continue;
		}
		else {
			Clear();
			//test();
			for (int i = 0; i < cmd.length(); i++) {
				if (cmd[i] == '(' || (cmd[i] <= 'Z' && cmd[i] >= 'A')) {
					pt.push(cmd[i]);
				}
				else {
					char c = pt.top(); pt.pop();
					char c2 = pt.top(); pt.pop();
					mt tmp = find(c), tmp2 = find(c2);
					if (tmp2->n != tmp->m) {
						cout << "error" << endl;
						flag = 1;
						break;
					}
					else {
						sum += (tmp2->m) * (tmp2->n) * (tmp->n);
						mt temp = new struct multi();
						temp->m = tmp2->m;
						temp->n = tmp->n;
						order += 1;
						temp->c = order;
						pt.pop();
						pt.push(order);
						Insert(temp);
					}
				}
			}
			if (!flag) {
				cout << sum << endl;
			}
			flag = 0;
		}
	}
	return 0;
}

写完后再看自己的代码是很粗糙的，比如给新的矩阵赋予新的字母，如果给出的表达式过长使得ASCll码超过了255该怎么办，还有每次给新的矩阵开辟新的空间插入链表中，计算下一个表达式的时候这些空间没有释放，还存留着，造成内存泄漏，是个很不好的习惯，而刘汝佳老师的代码很精简，比我的要好多了。。。

#include<iostream>
#include<cstdio>
#include<stack>
using namespace std;
struct multi
{
	int m;
	int n;
	multi(int a = 0, int b = 0) :m(a), n(b){}
} t[26];
stack<multi> p;

int main() {
	//freopen("D://in.txt", "r", stdin);
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		char c;
		cin >> c;
		cin >> t[c - 'A'].m >> t[c - 'A'].n;
	}
	string expr;
	while (cin >> expr) {
		bool error = false;
		int ans = 0;
		for (int i = 0; i < expr.length(); i++) {
			if (isalpha(expr[i])) {
				p.push(t[expr[i] - 'A']);
			}
			else if (expr[i] == ')') {
				multi t2 = p.top(); p.pop();
				multi t1 = p.top(); p.pop();
				if (t1.n != t2.m) {
					error = true;
					break;
				}
				else {
					ans += t1.m * t1.n * t2.n;
					p.push(multi(t1.m, t2.n));//直接创建一个multi结构体并入栈，这里第一次看过
				}
			}
		}
		if (error) {
			cout << "error\n";
		}
		else {
			cout << ans << endl;
		}
	}
	return 0;
}