Problem I
I Can Guess the Data Structure!
There is a bag-like data structure, supporting two operations:
1 x
Throw an element x into the bag.
2
Take out an element from the bag.
Given a sequence of operations with return values, you're going to guess the data structure. It is a stack (Last-In, First-Out), a queue (First-In, First-Out), a priority-queue (Always take out larger elements first) or something else that you can hardly imagine!
Input
There are several test cases. Each test case begins with a line containing a single integer n (1<=n<=1000). Each of the next n lines is either a type-1 command, or an integer 2 followed by an integer x. That means after executing a type-2 command, we get an element x without error. The value of x is always a positive integer not larger than 100. The input is terminated by end-of-file (EOF). The size of input file does not exceed 1MB.
Output
For each test case, output one of the following:
stack
It's definitely a stack.
queue
It's definitely a queue.
priority queue
It's definitely a priority queue.
impossible
It can't be a stack, a queue or a priority queue.
not sure
It can be more than one of the three data structures mentioned above.
Sample Input
6 1 1 1 2 1 3 2 1 2 2 2 3 6 1 1 1 2 1 3 2 3 2 2 2 1 2 1 1 2 2 4 1 2 1 1 2 1 2 2 7 1 2 1 5 1 1 1 3 2 5 1 4 2 4
Output for the Sample Input
queue not sure impossible stack priority queue
ＳＴＬ的应用，学习一下，做法就是先把输入的数据，存起来，然后用栈，队列，优先队列都试一下，看是不是满足就可以了。
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <stack> 5 #include <queue> 6 #include <iostream> 7 using namespace std; 8 struct data{ 9 int v; int k; 10 friend bool operator < (data a, data b){ 11 return a.v < b.v; 12 } 13 }s[1000+10]; 14 int n; 15 int cstack(){ 16 stack<int> qq; 17 for (int i = 0; i < n; ++i){ 18 if (s[i].k == 1) qq.push(s[i].v); 19 if (s[i].k == 2) { 20 if (qq.empty() == 1) return 0; 21 if (qq.top() == s[i].v) { 22 qq.pop(); continue; 23 } 24 else return 0; 25 } 26 } 27 return 1; 28 } 29 int cqueue(){ 30 queue<int> qq; 31 for (int i = 0; i < n; ++i){ 32 if (s[i].k == 1) qq.push(s[i].v); 33 if (s[i].k == 2){ 34 if (qq.empty() == 1) return 0; 35 if (qq.front() == s[i].v) { 36 qq.pop(); continue; 37 } 38 else return 0; 39 } 40 } 41 return 1; 42 } 43 int cpriqueue(){ 44 priority_queue<data> qq; 45 for (int i = 0; i < n; ++i){ 46 if (s[i].k == 1) qq.push(s[i]); 47 if (s[i].k == 2) { 48 if (qq.empty() == 1) return 0; 49 if (qq.top().v == s[i].v) { 50 qq.pop(); continue; 51 } 52 else return 0; 53 } 54 } 55 return 1; 56 } 57 int main(void){ 58 #ifndef ONLINE_JUDGE 59 freopen("11995.in", "r", stdin); 60 #endif 61 while (~scanf("%d", &n)){ 62 for (int i = 0; i < n; ++i){ 63 scanf("%d%d", &s[i].k, &s[i].v); 64 } 65 int st = cstack(), qu = cqueue(), pri = cpriqueue(); 66 if (st + qu + pri > 1) printf("not sure\n"); 67 else if (st) printf("stack\n"); 68 else if (qu) printf("queue\n"); 69 else if (pri) printf("priority queue\n"); 70 else printf("impossible\n"); 71 } 72 73 return 0; 74 }
关键是学会ＳＴＬ的使用，然后，还有结构体里面的友元函数，用来优先队列实现里面判断大小，注意，参数列表不能用 data &a, data &b会报错，看来Ｃ＋＋语法还得复习一下啊……