这道题比较简单。
需要注意的一些地方:
1、impossible: 所有的标记量都是false
2、not sure:同时存在2种情况或者同时存在三种情况.
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!
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.
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.
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
queue not sure impossible stack priority queue
Note: Please make sure to test your program with the gift I/O files before submitting!
直接对三种数据结构进行模拟,注意当前数据结构为空的时候的判断。
代码如下:
/*
* UVA_11995.cpp
*
* Created on: 2014年12月28日
* Author: Administrator
*/
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 1010;
int n;
int id[maxn];
int x[maxn];
bool isStack(){
stack s;
int i;
for(i = 0 ; i < n ; ++i){
if(id[i] == 1){
s.push(x[i]);
}else{
if(s.empty() != true){
int val = s.top();
s.pop();
if(x[i] != val){
return false;
}
}else{
return false;
}
}
}
return true;
}
bool isQueue(){
queue q;
int i;
for(i = 0 ; i < n ; ++i){
if(id[i] == 1){
q.push(x[i]);
}else{
if(q.empty() != true){
int val = q.front();
q.pop();
if(x[i] != val){
return false;
}
}else{
return false;
}
}
}
return true;
}
bool isPriority_Queue(){
priority_queue q;
int i;
for(i = 0 ; i < n ; ++i){
if(id[i] == 1){
q.push(x[i]);
}else{
if(q.empty() != true){
int val = q.top();
q.pop();
if(x[i] != val){
return false;
}
}else{
return false;
}
}
}
return true;
}
int main(){
while(scanf("%d",&n) != EOF){
memset(id,-1,sizeof(id));
memset(x,-1,sizeof(x));
int i;
for(i = 0 ; i < n ; ++i){
scanf("%d%d",&id[i],&x[i]);
}
bool st = isStack();
bool q = isQueue();
bool pq = isPriority_Queue();
if(st == false && q == false && pq == false){
printf("impossible\n");
}else if((st == true && q == true && pq == true) || (st == true && q == true && pq == false) || (st == true && q == false && pq == true) || (st == false && q == true && pq == true)){
printf("not sure\n");
}else if(st == true && q == false && pq == false){
printf("stack\n");
}else if(q == true && st == false && pq == false){
printf("queue\n");
}else if(pq == true && st == false && q == false){
printf("priority queue\n");
}
}
return 0;
}