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
Description
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.
Input
The last block consists of just one line containing 0.
Output
Sample Output
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 ",(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 "); else printf("No "); } } }
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 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 (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"
Output Specification
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.
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
/************************************************************************* > 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 ",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 ")) : (printf("%d ",ans)); return 0; }