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 并且括号中最多有两个 那么就非常easy了 遇到字母直接入栈 遇到反括号计算后入栈 然后就得到结果了
#include<cstdio>
#include<cctype>
#include<cstring>
using namespace std;
const int N = 1000;
int st[N], row[N], col[N], r[N], c[N]; int main()
{
int n, ans, top;
scanf("%d", &n);
char na[3], s[N];
for(int i = 1; i <= n; ++i)
{
scanf("%s", na);
int j = na[0] - 'A';
scanf("%d%d", &row[j], &col[j]);
} while(~scanf("%s", &s))
{
int i;
for(i = 0 ; i < 26; ++i)
c[i] = col[i], r[i] = row[i];
ans = top = 0; for(i = 0; s[i] != '\0'; ++i)
{
if(isalpha(s[i]))
{
int j = s[i] - 'A';
st[++top] = j;
} else if(s[i] == ')')
{
if(r[st[top]] != c[st[top - 1]]) break;
else
{
--top;
c[st[top]] = c[st[top + 1]];
ans += (r[st[top]] * c[st[top]] * r[st[top + 1]]);
}
}
}
if(s[i] == '\0') printf("%d\n", ans);
else printf("error\n");
}
return 0;
}
| Matrix Chain Multiplication |
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 Specification
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 (
),
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 Specification
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
UVa 442 Matrix Chain Multiplication(矩阵链,模拟栈)的更多相关文章
- 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(栈模拟水题+专治自闭)
题目: 给出一串表示矩阵相乘的字符串,问这字符串中的矩阵相乘中所有元素相乘的次数. 思路: 遍历字符串遇到字母将其表示的矩阵压入栈中,遇到‘)’就将栈中的两个矩阵弹出来,然后计算这两个矩阵的元素相乘的 ...
- UVa 442 Matrix Chain Multiplication(栈的应用)
题目链接: https://cn.vjudge.net/problem/UVA-442 /* 问题 输入有括号表示优先级的矩阵链乘式子,计算该式进行的乘法次数之和 解题思路 栈的应用,直接忽视左括号, ...
- 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 ...
- UVA442 Matrix Chain Multiplication 矩阵运算量计算(栈的简单应用)
栈的练习,如此水题竟然做了两个小时... 题意:给出矩阵大小和矩阵的运算顺序,判断能否相乘并求运算量. 我的算法很简单:比如(((((DE)F)G)H)I),遇到 (就cnt累计加一,字母入栈,遇到) ...
- 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 二十 Matrix Chain Multiplication
Matrix Chain Multiplication Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %l ...
- 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 ...
- ACM学习历程——UVA442 Matrix Chain Multiplication(栈)
Description Matrix Chain Multiplication Matrix Chain Multiplication Suppose you have to evaluate ...
随机推荐
- QT解析命令行(QCommandLineOption和QCommandLineParser类)
Qt从5.2版开始提供了两个类QCommandLineOption和QCommandLineParser来解析应用的命令行参数. 一.命令行写法命令行:"-abc" 在QComma ...
- eclipse如何查看类之间的引用关系
今天遇到这个问题:mark一点点: 在类名上单击右键.选择Reference->Workingspace快捷克债券Ctrl+Shift+G 版权声明:本文博客原创文章,博客,未经同意,不得转载.
- 14.5.2 Changing the Number or Size of InnoDB Redo Log Files 改变InnoDB Redo Log Files的数量
14.5.2 Changing the Number or Size of InnoDB Redo Log Files 改变InnoDB Redo Log Files的数量 改变InnoDB redo ...
- hdu 4284 Travel(floyd + TSP)
虽然题中有n<=100个点,但实际上你必须走过的点只有H<=15个.而且经过任意点但不消耗C[i]跟D[i]可以为无限次,所以可以floyd预处理出H个点的最短路,之后剩下的...就成了裸 ...
- BAD packet signature 18245 错误解决
1.错误信息 2014-7-15 2:46:38 org.apache.jk.common.MsgAjp processHeader 严重: BAD packet signature 18245 20 ...
- eclipse Maven构建的project无法公布lib到tomcat的解决方法
问题: eclipse导入基于Maven的web项目时,公布到tomcat中.发现lib文件夹及jar包没有公布过去. 解决方式: eclipse中,选择项目属性Properties --> D ...
- [Cocos2d-x]Android的android.mk文件通用版本
原文地址: http://blog.ready4go.com/blog/2013/10/12/update-android-dot-mk-with-local-src-files-and-local- ...
- Java NIO 完全学习笔记(转)
本篇博客依照 Java NIO Tutorial翻译,算是学习 Java NIO 的一个读书笔记.建议大家可以去阅读原文,相信你肯定会受益良多. 1. Java NIO Tutorial Java N ...
- Selenium来抓取动态加载的页面
一般的爬虫都是直接使用http协议,下载指定url的html内容,并对内容进行分析和抽取.在我写的爬虫框架webmagic里也使用了HttpClient来完成这样的任务. 但是有些页面是通过js以及a ...
- c 可变参数 定义可变参数的函数
定义可变参数的函数,需要在stdarg.h头文件中定义的va_list类型和va_start.va_arg.va_end三个宏. 定义可变参数函数 va_list ap; //实际是定义一个指针va ...