标签:eth expr pac next std ret its compute names
C++中提供了STL模板statck 在使用的时候更为方便
除了一般的队列外 还有STL更有双向队列可以使用 #include<deque> 声明:deque <type > name
应用举例1:铁轨问题
Yes No
Yes
/************************************************************************* > File Name: 6_2.cpp > Author:KID_XiaoYuan > Mail:kuailexy@126.com > Created Time: 2017年06月05日 星期一 18时41分08秒 > 算法描述:Rails :判断输入的数字能否通过栈的方式后输出 如果可以 输出YES否则输出NO > 样例输入: 5 1 2 3 4 5 3 4 5 2 1 > 样例输出:YES > 参考更高效代码:p141 ************************************************************************/ #include<stack> #include<iostream> #define MAX 100 using namespace std; int input[MAX]; int input2[MAX]; int main() { stack<int> s; int n,j = 0,k = 0 ; scanf("%d",&n); for(int i = 0; i < n; i++) { scanf("%d",&input[i]); } for(int i =0; i < n; i++) { scanf("%d",&input2[i]); } while(j < n) { s.push(input[j]); while(k < n && (s.top() == input2[k])) { s.pop(); k++; } j++; } printf("%s\n",(s.empty()?"YES":"NO")); return 0; }
改进版本:
#include<iostream> #include<cstdio> #include<cstring> #include<stack> #include<algorithm> using namespace std; int main() { int n; int target[2200]; while(~scanf("%d",&n))///输入也是个大问题啊。。 { if(n == 0) return 0; while(~scanf("%d",&target[1])) { if(target[1] == 0)///首先判断输入的第一个数 { puts("");///不要忘记换行. break; } for(int i = 2; i <= n; i++) scanf("%d",&target[i]); int a,b; a = b = 1; stack<int> s; bool mark = true; while(b <= n)///这就是判断是否符合出栈规则的核心; { if(a == target[b])///判断重位的元素; { a++; b++; } else if(!s.empty() && s.top() == target[b])///判断先进后出的规则,可以想象成倒叙。 { s.pop(); b++; } else if(a <= n)///之所以能这样,关键是因为入栈的顺序是连续的1~n数字; { s.push(a); a++; } else { mark = false; break; } } if(mark == true) printf("Yes\n"); else printf("No\n"); } } }
应用实例2:
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 arematrices. Since matrix multiplication is associative, the order in which multiplications areperformed is arbitrary. However, the number of elementary multiplications neededstrongly 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 multiplicationsneeded for a given evaluation strategy.
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 thename 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"
For each expression found in the second part of the input file, print one line containingthe 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.
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))
0 0 0 error 10000 error 3500 15000 40500 47500 15125
/************************************************************************* > File Name: 6_3.cpp > Author:KID_XiaoYuan > Mail:kuailexy@126.com > Created Time: 2017年06月05日 星期一 19时25分21秒 ************************************************************************/ #include<iostream> #include<string> #include<stack> using namespace std; typedef struct Data { public: char name; int a,b; /*void input (char &name ,int &a , int &b) { name = name; a = a; b = b; }*/ }data; int main() { data m[26]; stack<data> s; int n; cin>>n;//输入元素个数 for(int i = 0; i < n; i++)//输入元素信息 { char name; cin>>name; cin>>m[name- ‘A‘].a>>m[name-‘A‘].b; } bool error = false; string str;//输入计算字符串 cin>>str; int ans = 0; for(int i =0; i < str.length();i++) { if(isalpha(str[i]))//判断是否为字母 { s.push(m[ str[i] - ‘A‘] ); } else if( str[i] == ‘)‘)//如果是)则需要计算 { data m1, m2; m2 = s.top(); s.pop(); m1 = s.top(); s.pop(); if(m1.b != m2.a) { printf("ERROR DATA : M1.B = %d M2.A = %d\n",m1.b,m2.a); error = true; break; } ans += m1.a * m1.b * m2.b; data c; c.a = m1.a; c.b = m2.b; s.push(c); } } error ? (printf("error\n")) : (printf("%d\n",ans)); return 0; }
标签:eth expr pac next std ret its compute names
原文地址:http://www.cnblogs.com/KID-XiaoYuan/p/6947216.html