hdu1082 Matrix Chain Multiplication

Posted 2022-12-05 计算机科学家的世界

Problem Description Matrix multiplication problem is a typical example of dynamical programming. 

Suppose you have to evaluate an expression like A*B*C*D*E 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 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix.
There are two different strategies to compute A*B*C, namely (A*B)*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
  

 
Source University of Ulm Local Contest 1996  
Recommend We have carefully selected several similar problems for you:   1074  1081  1080  1505  1069    分析：        这里假设输入的括号是合法的，因此，程序里面不对输入匹配性做合法检验，程序里面主要检测矩阵相乘是否合法，比如，一个10X20的矩阵无法与30X10的矩阵相乘，所以，在此使用一个堆栈，从左到右分析输入串，碰到'('直接忽略，碰到')'则从栈里面弹出两个矩阵，再相乘（如果相乘不合法，则输出不合法，并返回），将相乘的结果压入堆栈，一直到扫描完输入串，此时再扫描堆栈。 代码如下所求：

#include <stdio.h>
#include <string.h>
#include <vector>

using namespace std;

typedef struct Record
	int m;
	int n;
Record;

Record r[100];
vector<Record> stk;

void Init();
void Caculate(char* buf);

int main()
	Init();
	char buf[10000];
	while(scanf("%s", buf) != EOF)
		Caculate(buf);
	


void Init()
	int len;
	scanf("%d", &len);
	char c;
	for(int i = 0; i < len; ++i)
		scanf("%c %c", &c, &c);
		c -= 'A';
		scanf("%d %d", &r[c].m, &r[c].n);
	


void Caculate(char* buf)
	int l = 0, r = 0, result = 0, len = strlen(buf);
	stk.clear();
	char c;
	for(int i = 0; i < len; ++i)
		c = buf[i];
		if(c == '(')
			++l;
		else if(c == ')')
			--l;
			if(stk.size() >= 2)
				Record lr = stk.back();
				stk.pop_back();
				Record ll = stk.back();
				stk.pop_back();
				if(lr.m != ll.n)
					printf("error\\n");
					return;
				else
					Record r;
					r.m = ll.m;
					r.n = lr.n;
					result += ll.m * ll.n * lr.n;
					stk.push_back(r);
				
			else if(stk.size() == 1)
				//do nothing
			else//0
				printf("error\\n");
				return;
			
		else//push stack
			stk.push_back(::r[c-'A']);
		
	
	Record lr;//= stk.back();
	//stk.pop_back();
	Record ll;// = stk.back();
	//stk.pop_back();
	Record rr;
	while(stk.size() > 1)
		lr = stk.back();
		stk.pop_back();
		ll = stk.back();
		stk.pop_back();
		if(ll.n != lr.m)
			printf("error\\n");
			return;
		else
			rr.m = ll.m;
			rr.n = lr.n;
			stk.push_back(rr);
			result += ll.m * ll.n * lr.n;
		
	
	printf("%d\\n", result);

提交结果如下所求：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
15596068	2015-11-21 23:45:33	Accepted	1082	0MS	1724K	1603 B	C++	BossJue