思路：即对数据结构进行模拟，若是其中一个，则输出对应数据结构，若可能是其中的两个，则输出not sure 若都不是则输出impossible。
代码：
#include<stdio.h>
#include<stack>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
//0: queue 1:stack 2:Priority Queue
int type[3];
//操作结构 保存操作的类型与操作数
struct oper {
int op, d;
}o[1005];
int main() {
int n;
while (~scanf("%d", &n)){
memset(type, 0, sizeof type);
for (int i = 0; i < n; i++)
scanf("%d%d", &o[i].op, &o[i].d);
queue<int>q;
stack<int>s;
priority_queue<int>pq;
for (int i = 0; i < n; i++) {
if (o[i].op == 1) {
q.push(o[i].d);
}
else {
if (q.empty()) break;
int e = q.front();
q.pop();
if (o[i].d != e) break;
}
if (i == n - 1) type[0] = 1;
}
for (int i = 0; i < n; i++) {
if (o[i].op == 1)
s.push(o[i].d);
else {
if (s.empty()) break;
int e = s.top();
s.pop();
if (e != o[i].d) break;
}
if (i == n - 1) type[1] = 1;
}
for (int i = 0; i < n; i++) {
if (o[i].op == 1)
pq.push(o[i].d);
else {
if (pq.empty()) break;
int e = pq.top();
pq.pop();
if (e != o[i].d) break;
}
if (i == n - 1) type[2] = 1;
}
int sum = 0;
for (int i = 0; i < 3; i++)
sum += type[i];
if (sum == 0)
printf("impossible\n");
else if (sum > 1)
printf("not sure\n");
else if (type[0] == 1)
printf("queue\n");
else if (type[1] == 1)
printf("stack\n");
else if (type[2] == 1)
printf("priority queue\n");
}
return 0;
}