学步园

这道题就是典型的有优先队列

以不一样的标准排序，还有经常性了插入和删除

如果是优先队列，就是按照x由大到小排序；队列，按标号由小到大；栈，按标号有大到小。

注意：一定要考虑到当出栈大于入栈的情况，这样是impossible，而只有入栈没有出栈是not sure；入栈大于出栈或者等于出栈都是正常处理的

考虑周全

代码：

#include <queue>
#include <cstdio>
using namespace std;

struct pri_q {
    int x;
    bool operator < ( const pri_q& a) const {
        return x < a.x;
    }
}pri;
struct stac {
    int x, i;
    bool operator < ( const stac& a ) const {
        return i < a.i;
    }
}sta;
struct que {
    int x, i;
    bool operator < ( const que& a ) const {
        return i > a.i;
    }
}qu;
bool vis[3];
int main() 
{
    int n, op, num, ans, x1, x2;
    while ( scanf("%d", &n) != EOF ) {
        ans = 3, x1 = x2 = 0;
        priority_queue <pri_q> pq1;
        priority_queue <stac> pq2;
        priority_queue <que> pq3;
        vis[0] = vis[1] = vis[2] = 0;
        for ( int i = 0; i < n; ++i ) {
            scanf("%d%d", &op, &num);
            if ( op == 1 ) {
                x1++;
                if ( !vis[0] ) pri.x = num, pq1.push(pri);
                if ( !vis[1] ) sta.x = num, sta.i = i, pq2.push(sta);
                if ( !vis[2] ) qu.x = num, qu.i = i, pq3.push(qu);
            }
            else if ( op == 2 ) {
                x2++;
                if ( !pq1.empty() ) {
                    pri = pq1.top(); pq1.pop(); 
                    if ( pri.x != num ) vis[0] = true;
                }
                if ( !pq2.empty() ) {
                    sta = pq2.top(); pq2.pop(); 
                    if ( sta.x != num ) vis[1] = true;
                }
                if ( !pq3.empty() ) {
                    qu = pq3.top(); pq3.pop(); 
                    if ( qu.x != num ) vis[2] = true;
                }
            }
        }
        for ( int i = 0; i < 3; ++i ) if ( vis[i] ) ans--;
        if ( ans < 1 || x2 > x1 ) printf("impossible\n");
        else if ( ans > 1 ) printf("not sure\n");
        else if ( !vis[0] ) printf("priority queue\n");
        else if ( !vis[1] ) printf("stack\n");
        else if ( !vis[2] ) printf("queue\n");
    }
}