UVa 442 Matrix Chain Multiplication(矩阵链,模拟栈)
2015-12-13 21:23
513 查看
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
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.
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):
Otherwise print one line containing the number of elementary multiplications needed to evaluate the expression in the way specified by the parentheses.
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 , row , col , r , c ; int main() { int n, ans, top; scanf("%d", &n); char na[3], s ; 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 |
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
相关文章推荐
- Determining IP information for eth1... failed; no link present. Check cable? 解决办法
- Determining IP information for eth1... failed; no link present. Check cable? 解决办法
- MINI2440移植xenomai记录
- 机器学习: 朴素贝叶斯(Naive Bayes)
- LeetCode OJ——Submission Details
- Chain of Responsibility(职责链设计模式)
- 【转载】关于RAID 1+0和RAID 0+1的比较
- 从Container内存监控限制到CPU使用率限制方案
- linker command failed with exit code 1 (use -v to see invocation)
- 缩略图Thumbnails
- AIDL
- 静态局部变量 http://baike.baidu.com/link?url=h5FJNxRXfawWPNdJEzqWHpKN1HMk6u8wXLYgg8VYCqgd8MbypeKVeaOgZB0B-
- U盘安装Ubuntu——关于UltraISO打开Ubuntu只有EFI文件夹的解决方法
- VM虚拟机 安装OS X 错误vcpu-0:VERIFY vmcore/vmm/main/physMem_monitor.c:1123
- VM虚拟机 安装OS X 错误vcpu-0:VERIFY vmcore/vmm/main/physMem_monitor.c:1123
- AI
- org.hibernate.exception.ConstraintViolationException: could not insert:
- 安卓错误总结:Invalid file name: must contain only [a-z0-9_.]
- Lobes of the brain
- malloc()和calloc()区别 以及memset() http://blog.csdn.net/ermuzhi/article/details/7833701