UVA11995 优先队列

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

61 11 21 32 12 22 361 11 21 32 32 22 121 12 241 21 12 12 271 21 51 11 32 51 42 4

Output for the Sample Input

queuenot sureimpossiblestackpriority queue

遇到的问题和解题思路：

       首先看到这道题以为是单纯的水题，然后写了2个小时没写出来。看了一下别人写的代码以后发现，原来是理解出错了TAT。

       这道题的我理解错的地方在于，就是你输入进去的东西是一个一个的读入的，而不是全部读入了以后再比较，所以说TAT我就错在这里了。

给出AC代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<stack>using namespace std;int n, st, qu, pr;int a[1005][2];queue<int> que;stack<int> s;priority_queue<int> pq;void solve(int x ){if(a[x][0] == 1){que.push(a[x][1]);s.push(a[x][1]);pq.push(a[x][1]);}else {if(que.empty())qu = 0;else{if(que.front() != a[x][1]){que.pop();qu = 0;}else que.pop();}if(s.empty())st = 0;else {if(s.top() != a[x][1]){s.pop();st = 0;}else s.pop();}if(pq.empty())pr = 0;else{if(pq.top() != a[x][1]){pq.pop();pr = 0;}else pq.pop();}}}int main(){while(scanf("%d", &n)!=EOF){int i, j;st = qu = pr = 1;while(!que.empty())que.pop();while(!s.empty())s.pop();while(!pq.empty())pq.pop();for (i = 0; i < n; i++){for (j = 0; j < 2; j++){scanf("%d",&a[i][j]);}solve (i );//printf("%d %d %d\n", st, qu, pr);}if(st + qu + pr > 1)printf("not sure\n");else if(st + qu + pr == 0)printf("impossible\n");else{if(st == 1)printf("stack\n");else if(qu == 1)printf("queue\n");else if(pr == 1)printf("priority queue\n");}}return 0;}

给出测试数据，基本上所有讨论的种类都有了：（应该吧）

6
1 1
1 2
1 3
1 4
1 5
2 1
6
1 1
1 2
1 3
1 4
1 5
2 5
6
1 5
1 4
1 3
1 2
1 1
2 5
6
1 5
1 4
1 3
1 2
1 1
2 1
5
1 3
1 2
2 3
2 2
2 1
4
1 3
2 3
2 2
1 2/*最后一种，就是impossible，因为先是一个1，所以输入了3，后来说连着消除两次，显然这个是不可能的事情，所以是impossible*/

之前写的代码其中有两个问题，其中一个就是在每个函数中的x和y的值得大小不确定，当y>x的时候，就会出现RE。还有一个就是之前说的，题目理解错误

下面给出错误的代码，来警示错误：

#include<cstdio>#include<cstring>#include<queue>#include<stack>#include<algorithm>using namespace std;int n;int a[1005], b[1005];int ty[5];int pend, x, y;bool queue1(){queue<int> que;for(int i = 0; i < x; i++)    que.push(a[i]);    for(int i = 0; i < y; i++){    int d = que.front();    que.pop();    if(d != b[i])return false;    }    ty[2] = 1;    return true;}bool stack1(){stack<int> s;for(int i = 0; i < x; i++) s.push(a[i]);for(int i = 0; i < y; i++){int d = s.top();s.pop();if(d != b[i])return false;}ty[1] = 1;return true;}bool priority1(){priority_queue<int> pq;for(int i = 0; i < x; i++)pq.push(a[i]);for(int i = 0; i < y; i++){int d = pq.top();pq.pop();if(d != b[i])return false;}ty[3] = 1;return true;}int main(){while(scanf("%d", &n)!=EOF){x = 0, y = 0;memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));memset(ty, 0, sizeof(ty));for(int i = 0; i < n; i++){scanf("%d",&pend);if(pend == 1)scanf("%d",&a[x++]);if(pend == 2)scanf("%d",&b[y++]);}int res = 0;res += queue1();res += stack1();res +=priority1();if(res >= 2)printf("not sure\n");else if(res == 0)printf("impossible\n");else if(res == 1){if(ty[1])printf("stack\n");else if(ty[2])printf("queue\n");else if(ty[3])printf("priority queue\n");}}return 0;}