Matrix Chain Multiplication UVA - 442
Suppose you have to evaluate an expression like ABCDE where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.
For example, let A be a 5010 matrix, B a 1020 matrix and C a 205 matrix. There are two different strategies to compute ABC, namely (AB)C and A(B*C).
The first one takes 15000 elementary multiplications, but the second one only 3500.
Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy.
Input
Input consists of two parts: a list of matrices and a list of expressions.
The first line of the input file contains one integer n (1 ≤ n ≤ 26), representing the number of matrices in the first part. The next n lines each contain one capital letter, specifying the name of the matrix, and two integers, specifying the number of rows and columns of the matrix.
The second part of the input file strictly adheres to the following syntax (given in EBNF):
SecondPart = Line { Line } < EOF>
Line = Expression < CR>
Expression = Matrix | "(" Expression Expression ")"
Matrix = "A" | "B" | "C" | ... | "X" | "Y" | "Z"
Output
For each expression found in the second part of the input file, print one line containing the word ‘error’ if evaluation of the expression leads to an error due to non-matching matrices. Otherwise print one line containing the number of elementary multiplications needed to evaluate the expression in the way specified by the parentheses.
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
HINT
使用map来记录矩阵,使用栈来存入数组。思路很简单,直接看代码就好。
注意:每一次做乘法计数的时候偶要判断是不是字母,否则会出错!!!
Accepted
#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
using namespace std;
int main()
{
map<string, vector<int>>M;
long long int sum = 0;
int n,a,b;
string t,s;
cin >> n;
while (n--) {
cin >> t >> a >> b;
M[t].push_back(a);
M[t].push_back(b);
}
getchar(); //
while (getline(cin, s)) {
stack<int>list;
sum = 0;
if (s.length() == 1)cout << 0 << endl;
else {
for (int i = 0;i < s.length();i++) {
if (s[i] == '(')continue;
if (s[i] == ')') { //计算
b = list.top();list.pop();
a = list.top();list.pop();
if (a != list.top()) { cout << "error" << endl;sum = -1;break; }
list.pop();
sum += a * b * list.top();
list.push(b);
}
else { //入栈
t = s[i];
list.push(M[t][0]);
list.push(M[t][1]);
}
}
if (sum != -1) cout << sum << endl;
}
}
}
Matrix Chain Multiplication UVA - 442的更多相关文章
- UVA 442 二十 Matrix Chain Multiplication
Matrix Chain Multiplication Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %l ...
- 例题6-3 Matrix Chain Multiplication ,Uva 442
这个题思路没有任何问题,但还是做了近三个小时,其中2个多小时调试 得到的经验有以下几点: 一定学会调试,掌握输出中间量的技巧,加强gdb调试的学习 有时候代码不对,得到的结果却是对的(之后总结以下常见 ...
- UVA——442 Matrix Chain Multiplication
442 Matrix Chain MultiplicationSuppose you have to evaluate an expression like A*B*C*D*E where A,B,C ...
- UVa 442 Matrix Chain Multiplication(矩阵链,模拟栈)
意甲冠军 由于矩阵乘法计算链表达的数量,需要的计算 后的电流等于行的矩阵的矩阵的列数 他们乘足够的人才 非法输出error 输入是严格合法的 即使仅仅有两个相乘也会用括号括起来 并且括号中 ...
- Matrix Chain Multiplication[HDU1082]
Matrix Chain Multiplication Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (J ...
- UVa442 Matrix Chain Multiplication
// UVa442 Matrix Chain Multiplication // 题意:输入n个矩阵的维度和一些矩阵链乘表达式,输出乘法的次数.假定A和m*n的,B是n*p的,那么AB是m*p的,乘法 ...
- Matrix Chain Multiplication(表达式求值用栈操作)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1082 Matrix Chain Multiplication Time Limit: 2000/100 ...
- ACM学习历程——UVA442 Matrix Chain Multiplication(栈)
Description Matrix Chain Multiplication Matrix Chain Multiplication Suppose you have to evaluate ...
- UVa 442 (栈) Matrix Chain Multiplication
题意: 给出一个矩阵表达式,计算总的乘法次数. 分析: 基本的数学知识:一个m×n的矩阵A和n×s的矩阵B,计算AB的乘法次数为m×n×s.只有A的列数和B的行数相等时,两个矩阵才能进行乘法运算. 表 ...
随机推荐
- python进阶(1)Lambda表达式
Lambda表达式 lambda表示的是匿名函数,不需要用def来声明,一句话就可以声明出一个函数 语法 函数名 = lambda 参数:返回值 注意点 1.函数的参数可以有多个,多个参数之间用逗号隔 ...
- 配置JDK环境及其相关问题
1.首先找到JDK的安装目录 如果忘记了安装目录在那个地方,可以通过dos命令java -verbose,进行查看 配置jdk环境 新建系统变量JAVA_HOME: 编辑系统变量Path: 新建系统变 ...
- Vue学习笔记-vue调试工具vue-devtools安装及使用
一 使用环境: windows 7 64位操作系统 二 vue调试工具vue-devtools安装及使用 1.下载: 百度中查找 "vue-devtools下载" 找到最新 ...
- 小白养成记——MySQL中的排名函数
1.ROW_NUMBER() 函数 依次排序,没有并列名次.如 SELECT st.ID '学号', st.`NAME` '姓名', sc.SCORE '成绩', ROW_NUMBER() OVER( ...
- HashSet为什么可以有序输出?
首先HashSet是不保证有序,而不是保证无序,因为在HashSet中,元素是按照他们的hashCode值排序存储的.对于单个字符而言,这些hashCode就是ASCII码,因此,当按顺序添加自然数或 ...
- void指针及指针的多次赋值的理解
1.void指针的类型转换 int A::functionCommamd(const DWORD _from,const DWORD _to,const DWORD Event_type,void * ...
- Hi3519 SDK搭建、问题总结及yolov3 RFCN的运行结果与测试
下面记录一下,在搭建Hi3519A SDK的注意事项与遇到的问题解决,及Hi3519A SDK环境下进行yolov3.RFCN的测试.(具体的Hi3519A的SDK环境搭建参考后面随笔-Hi3559A ...
- Node更丝滑的打开方式
Node更丝滑的打开方式 1. 使用背景 最近前端的一个项目,使用gulp作为工程化.在运行过程中出现如下错误 gulp[3192]: src\node_contextify.cc:628: Asse ...
- CCF(地铁修建):向前星+dijikstra+求a到b所有路径中最长边中的最小值
地铁修建 201703-4 这题就是最短路的一种变形,不是求两点之间的最短路,而是求所有路径中的最长边的最小值. 这里还是使用d数组,但是定义不同了,这里的d[i]就是表示从起点到i的路径中最长边中的 ...
- c# float类型和double类型相乘出现精度丢失
c# float类型和double类型相乘出现精度丢失 double db = 4.0; double db2 = 1.3; float f = 1.3F; float f2 = 4.0F; Deci ...