I Can Guess the Data Structure!
UVA - 11995題目鏈接：https://vjudge.net/problem/UVA-11995
題意：給定N個操作，詢問這些操作能否滿足棧、隊列、優先隊列。
題解：建立棧、隊列、優先隊列，判斷操作是否合法即可，操作可能有錯，所以pop需要判斷empty。
AC代碼：
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <cstdio>
using namespace std;
#define _for(i,a,b) for(int i=a;i<=b;i++)
stack<int> S;
queue<int> Q;
priority_queue<int> P;
int t,n,ans[4];
void intc()
{
while(Q.size())Q.pop();
while(P.size())P.pop();
while(S.size())S.pop();
_for(i,1,3)ans[i]=0;
}
int main(int argc, char const *argv[])
{
while(scanf("%d",&n)!=EOF)
{
intc();
_for(i,1,n)
{
int x,y;
cin>>x>>y;
if(x==1)
{
S.push(y);
P.push(y);
Q.push(y);
}
else
{
if(!S.empty())
{
int now = S.top();
S.pop();
if(now!=y)ans[1]=1;
}
else ans[1]=1;
if(!Q.empty())
{
int now = Q.front();
if(now!=y)ans[2]=1;
Q.pop();
}
else ans[2]=1;
if(!P.empty())
{
int now=P.top();
if(now!=y)ans[3]=1;
P.pop();
}
else ans[3]=1;
}
}
int h=0;
_for(i,1,3)if(ans[i]==1)h++;
if(h<2)cout<<"not sure"<<endl;
else if(h==3)cout<<"impossible"<<endl;
else
{
if(ans[1]==0)cout<<"stack"<<endl;
if(ans[2]==0)cout<<"queue"<<endl;
if(ans[3]==0)cout<<"priority queue"<<endl;
}
}
return 0;
}