zoj 1094 Matrix Chain Multiplication 栈应用

//利用stack计算表达式
//类似计算中序表达式那样，用两个栈，一个存字母，一个存符号
//这题要自己存入*号，要用一个变量储存之前输入的符号，判断是否要把*入栈
//如果前一个符号是（，则不需要。
#include <iostream>
#include <stack>
#include <cstdio>
using namespace std;

struct info
{
int row, col;
info() {}
info (int row, int col)
{
this->row = row;
this->col = col;
}
};

const int N = 30;
stack<char> oprator;
stack<info> matrix;

info a
;
int n;

int main()
{
char c, s, last;
int ans = 0;
info num1, num2;
bool isError;
while (cin >> n)
{
for (int i = 0; i < n; i++)
cin >> c >> a[i].row >> a[i].col;

isError = false;
getchar();
last = '(';
while (scanf("%c", &c) != EOF)
{
if (c == '\n')
{
if (isError)
cout << "error" << endl;
else
cout << ans << endl;
while (!matrix.empty())
matrix.pop();
while (!oprator.empty())
oprator.pop();
isError = false;
ans = 0;
last = '(';
continue;
}
if (isError)
continue;
if (c == '(')
{
if (last != c)
oprator.push('*');
oprator.push(c);
last = c;
}
else if (c == ')')
{
s = oprator.top();
oprator.pop();
while (s != '(')
{
num1 = matrix.top();
matrix.pop();
num2 = matrix.top();
matrix.pop();
if (num2.col != num1.row)
isError = true;
else
{
ans += num2.row*num2.col*num1.col;
matrix.push(info(num2.row, num1.col));
}
s = oprator.top();
oprator.pop();
}
last = '0';
}
else
{
if (last != '(')
oprator.push('*');
matrix.push(info(a[c-'A'].row, a[c-'A'].col));
last = c;
}
}
}
return 0;
}