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首先学习队列，队列有先进先出的特性。广度优先遍历需要基于队列实现，C++中的stl引入了队列的实现方式。队列支持push()，进入队尾，pop()出队，队头出队，front()获取队首元素，back()获取队尾元素，empty()判断队列是否为空。
第一题是约瑟夫环问题。
#include <stdio.h>
#include <queue>
using namespace std;
int main()
{queue<int> myqueue;int n,p,m;while(scanf("%d%d%d", &n,&p,&m)!=EOF){if(n ==0&&p==0&&m==0) break;//生成第一轮报数的队列//p, p+1,p+2,...,n,1,2,....int no = p;//孩子编号for(int i = 0; i < n;i++){myqueue.push(no);no++;if(no>n) no=1;}//开始报数int say = 1;while(1){int cur = myqueue.front();myqueue.pop();if(say == m){say = 1;if(myqueue.empty()){printf("%d\n", cur);break;}else{printf("%d,", cur);}}else {say++;myqueue.push(cur);}}}return 0;
}第二题是排队打饭，分情况讨论即可，主要注意time是long long类型，使用int这题过不了。
#include <stdio.h>
#include <queue>
using namespace std;
struct Student{int a;//到达时刻int t;//打饭耗时int b;//最大等待时间
};
int main(){queue<Student> que;int n;scanf("%d", &n);for(int i = 0; i<n;i++){int ai,ti,bi;scanf("%d%d%d", &ai, &ti,&bi);Student s1;s1.a = ai;s1.b = bi;s1.t = ti;que.push(s1);//读入学生序列}long long time = 1;//记录当前时刻while(!que.empty()){Student stu = que.front();if(time>(stu.a+stu.b)){printf("-1 ");que.pop();}else if(time>=stu.a&&time<=(stu.a+stu.b)){printf("%d ", time);time += stu.t;que.pop();}else if(time<stu.a){time = stu.a;printf("%lld ", time);time += stu.t;que.pop();}}
}下面学习栈，后进先出。深度优先遍历、表达式解析、递归会使用栈。stack支持push()入站，pop()出栈，top()获取栈顶元素，栈只支持获取栈顶元素，empty()获取栈是否为空。
第三题是编排字符串。栈的简单应用。
#include <stdio.h>
#include <stack>
#include <string>
using namespace std;
int main(){int n;scanf("%d", &n);stack <string> st1;for(int i = 0; i<n;i++){char mid[20];scanf("%s", mid);string str = mid;st1.push(str);stack<string> mystack;mystack = st1;int num = 1;while(!mystack.empty()){if(num>4) break;string str1 = mystack.top();printf("%d=%s ", num, str1.c_str());num++;mystack.pop();}printf("\n");}}第四题是括号匹配，经典题目。
#include <stdio.h>
#include <stack>
#include <string>
using namespace std;
int main()
{char arr[20000];scanf("%s", arr);string str = arr;stack<char> mystack;mystack.push(str[0]);for(int i = 1;i < str.size();i++){if(str[i] == '<'||str[i] == '('||str[i] == '{'||str[i] == '['){mystack.push(str[i]);}else if(str[i]=='>'){if(mystack.empty()) {printf("no");return 0;}char res = mystack.top();if(res == '<') mystack.pop();else {printf("no");return 0;}}else if(str[i]==')'){if(mystack.empty()) {printf("no");return 0;}char res = mystack.top();if(res == '(') mystack.pop();else {printf("no");return 0;}}else if(str[i]=='}'){if(mystack.empty()) {printf("no");return 0;}char res = mystack.top();if(res == '{') mystack.pop();else {printf("no");return 0;}}else if(str[i]==']'){if(mystack.empty()) {printf("no");return 0;}char res = mystack.top();if(res == '[') mystack.pop();else {printf("no");return 0;}}}if(mystack.empty()==true) printf("yes");else printf("no");return 0;
}第五题是堆栈的使用，简单模拟。
#include <stdio.h>
#include <string>
#include <stack>
using namespace std;
int main() {int n;while (scanf("%d", &n) != EOF) {stack <int> mystack;for (int i = 0; i < n; i++) {char op[2];scanf("%s", op);string op2=op;if (op2 == "A") {if (mystack.empty()) printf("E\n");else printf("%d\n", mystack.top());} else if (op2 == "O") {if (!mystack.empty()) mystack.pop();} else if (op2 == "P") {int num;scanf("%d", &num);mystack.push(num);}}}return 0;
}第六题是计算表达式，写了半天代码逻辑不对，气完了。建立运算符栈与运算数栈。从左向右扫描表达式，遇到操作数就加入操作数栈，扫描到运算符，若运算符栈为空，则直接压入运算符栈，若运算符栈不为空但当前运算符优先级大于栈顶运算符，也执行压栈操作。若运算符栈不为空但当前运算符优先级低于栈顶运算符，则弹出栈内的运算符，同时弹出操作数，直到当前运算符优先级再次大于栈顶运算符，压入栈中。
#include <stdio.h>
#include <string>
#include <stack>
#include <map>
using namespace std;
int main() {char str[1000] = {0};map<char, int> prio = {{'\0', 0}, {'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}};while (scanf("%s", str) != EOF) {string numstr = "";stack <char> opstack;stack <double> numstack;for (int i = 0;; i++) {if (str[i] >= '0' && str[i] <= '9') {numstr.push_back(str[i]);} else {double num = stod(numstr);numstr = "";//数值提取numstack.push(num);//什么时候弹栈？栈非空&&新的op的优先级不高于栈顶的优先级//循环弹栈和计算while (!opstack.empty() && prio[str[i]] <= prio[opstack.top()]) {double right = numstack.top();numstack.pop();double left = numstack.top();numstack.pop();char curop = opstack.top();opstack.pop();if (curop == '+') numstack.push(left + right);else if (curop == '-') numstack.push(left - right);else if (curop == '*') numstack.push(left * right);else if (curop == '/') numstack.push(left / right);}//栈为空或者新运算符的优先级大于栈顶运算符if (str[i] == '\0') {printf("%d\n", (int)numstack.top());break;} else opstack.push(str[i]);}}}return 0;
}第七题是模拟出入栈。
#include <bits/stdc++.h>
using namespace std;int main() {string s;while (cin >> s) {stack<char> stk;int j = 0; //j用来扫描出栈序列s的for (char i = 'a'; i <= 'z'; ++ i) {stk.push(i); //每次按顺序入栈一个while (stk.size() && stk.top() == s[j]) {j++; //出栈序列向后走匹配下一个stk.pop(); //出栈}}if (stk.empty()) cout << "yes\n"; //栈空了，就是匹配的else cout << "no\n";}return 0;
}