Dima and Containers CodeForces - 358C

原题链接
考察:模拟,贪心
思路:
  万恶的模拟+离线处理题.
  对于相邻的0之间的数字,最大的放在stack里,其次在队列,最后在双端队列.
  要注意的是可能末尾没有0

Code

#include <iostream> 
#include <cstring>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <algorithm>
using namespace std;
const int N = 100010;
int n,op,a[N];
string ans[N];
struct Node{
	int id,val;
	bool operator<(const Node& x){
		return this->val<x.val;
	}
}node[N];
void Myprintf(int cnt,int pos)
{
	if(!cnt) {puts("0");return;}
	sort(node+1,node+cnt+1);
	if(cnt==1)
	{
		puts("pushStack");
		if(!pos) puts("1 popStack");
		return;
	}
	if(cnt==2) 
	{
		printf("pushStack\npushQueue\n");
		if(!pos) printf("2 popStack popQueue\n");
		return;
	}
	ans[node[cnt].id] = "pushStack";
	ans[node[cnt-1].id] = "pushQueue";
	ans[node[cnt-2].id] = "pushFront";
	for(int i=cnt-3;i>=1;i--) ans[node[i].id] = "pushBack";
	for(int i=1;i<=cnt;i++) printf("%s\n",ans[i].c_str());
	if(!pos) printf("3 popStack popQueue popFront\n");
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	int cnt = 0;
	for(int i=1;i<=n;i++)
	{
		if(a[i])
		{
			cnt++;
			node[cnt].id = cnt;
			node[cnt].val = a[i];
		}else{
			Myprintf(cnt,a[i]);
			cnt = 0;
		}
	}
	if(cnt) Myprintf(cnt,1);
	return 0;
}