stack UVA 442 Matrix Chain Multiplication

题目传送门

题意：给出每个矩阵的行列，计算矩阵的表达式，如果错误输出error，否则输出答案

分析：表达式求值，stack 容器的应用：矩阵的表达式求值A 矩阵是a * b，B 矩阵是b * c，则A * B 是a * c。遇到')'弹出两个矩阵相乘，错误的话直接break

收获：以前做过了，现在会表达式求值后，这题也太容易了

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-29 10:22:51
* File Name     :UVA_442.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Martrix	{
	int a, b;
	Martrix (int _a = 0, int _b = 0) : a (_a), b (_b) {};
}m[26];

int main(void)    {
	int n;
	string name;
	cin >> n;
	for (int i=1; i<=n; ++i)	{
		cin >> name;	int k = name[0] - 'A';
		cin >> m[k].a >> m[k].b;
	}

	string exp;
	stack<Martrix> S;
	while (cin >> exp)	{
        bool error = false;
        int ans = 0;
		int len = exp.length ();
		for (int i=0; i<len; ++i)	{
			if ('A' <= exp[i] && exp[i] <= 'Z')	S.push (m[exp[i]-'A']);
			else if (exp[i] == ')')	{
				Martrix m1 = S.top ();	S.pop ();
				Martrix m2 = S.top ();	S.pop ();
				if (m2.b != m1.a)	{
					error = true;	break;
				}
				ans += m2.a * m2.b * m1.b;
				S.push (Martrix (m2.a, m1.b));
			}
		}
		if (error)	cout << "error" << endl;
		else	cout << ans << endl;
	}

    return 0;
}