B:I Can Guess the Data Structure!（建议使用栈、队列和优先队列来模拟）
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
题意：这道题就是通过输入若干数据结构类型，让你来判断，理解题意后并不难，但在规定时间并没有做出来，在补题时，通过学长给的建议有了点思路
首先开始我的代码如下：很遗憾结果时间超限
#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n))
{
stack<int>x;
queue<int>y;
priority_queue<int>z;
int p,q,i,j,f[3];
for(i=0;i<3;i++)
{
f[i]=1;
}
for(i=0;i<n;i++)
{
scanf("%d %d",&p,&q);
if(p==1)
{
x.push(q);
y.push(q);
z.push(q);
}
else
{
if(!x.empty())
{
if(x.top()!=q) //多次判断导致时间超限
{
f[0]=0;
}
x.pop();
}
else
f[0]=0;
if(!y.empty())
{
if(y.front()!=q) //多次判断导致时间超限
f[1]=0;
y.pop();
}
else
f[1]=0;
if(!z.empty())
{
if(z.top()!=q) //多次判断导致时间超限
f[2]=0;
z.pop();
}
else
f[2]=0;
}
}
int sum=0;
for(i=0;i<3;i++)
{
if(f[i]==1)
sum++;
}
if(sum==0)
printf("impossible\n");
else if(sum>1)
printf("not sure\n");
else
{
if(f[0]==1)
printf("stack\n");
if(f[1]==1)
printf("queue\n");
if(f[2]==1)
printf("priority queue\n");
}
}
return 0;
}
但在后面自己通过不断修改和寻问学长，解决了这个问题，最后***AC***代码如下：
#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
using namespace std;
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
stack<int>x;
queue<int>y;
priority_queue<int>z;
int f[3];
for(i=0;i<3;i++)
{
f[i]=1;
}
for(i=0;i<n;i++)
{
int p,q;
scanf("%d%d",&p,&q);
if(p==1)
{
x.push(q);
y.push(q);
z.push(q);
}
else
{
if(!x.empty())
{
int a=x.top();
if(a!=q)
{
f[0]=0;
}
x.pop();
}
else
f[0]=0;
if(!y.empty())
{
int b=y.front();
if(b!=q)
f[1]=0;
y.pop();
}
else
f[1]=0;
if(!z.empty())
{
int c=z.top();
if(c!=q)
f[2]=0;
z.pop();
}
else
f[2]=0;
}
}
int sum=0;
for(i=0;i<3;i++)
{
if(f[i]==1)
sum++;
}
if(sum==0)
printf("impossible\n");
else if(sum>1)
printf("not sure\n");
else
{
if(f[0]==1)
printf("stack\n");
else if(f[1]==1)
printf("queue\n");
else if(f[2]==1)
printf("priority queue\n");
}
}
return 0;
}
小白还在不断努力