The SetStack Computer UVA

大小为n,m的set调用stl方法的复杂度是O(n+m),听说常数大(?)

这题的集合中，元素还是集合，

如果我们把每个集合都分配一个编号，那么就很容易用set表达出集合的包含关系

这个实现需要集合查找编码与编码查找集合

具体看代码吧

#include<bits/stdc++.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<iostream>
#include<math.h>
using namespace std;
#define ll long long
const int maxn=1e4+7;
const int inf=0x3f3f3f3f;
#define FOR(n) for(int i=1;i<=n;i++)
#define pb push_back

/******************************************************/
namespace fastIO{  
	#define BUF_SIZE 100000  
	#define OUT_SIZE 100000  
	#define ll long long  
	//fread->read  
	bool IOerror=0;  
	inline char nc(){  
		static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;  
		if (p1==pend){  
			p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);  
		if (pend==p1){IOerror=1;return -1;}  
		//{printf("IO error!
");system("pause");for (;;);exit(0);}  
	}  
	return *p1++;  
}  
inline bool blank(char ch){return ch==' '||ch=='
'||ch=='
'||ch=='	';}  
inline void read(int &x){  
	bool sign=0; char ch=nc(); x=0;  
	for (;blank(ch);ch=nc());  
	if (IOerror)return;  
	if (ch=='-')sign=1,ch=nc();  
	for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';  
	if (sign)x=-x;  
}  
inline void read(ll &x){  
	bool sign=0; char ch=nc(); x=0;  
	for (;blank(ch);ch=nc());  
	if (IOerror)return;  
	if (ch=='-')sign=1,ch=nc();  
	for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';  
	if (sign)x=-x;  
}  
inline void read(double &x){  
	bool sign=0; char ch=nc(); x=0;  
	for (;blank(ch);ch=nc());  
	if (IOerror)return;  
	if (ch=='-')sign=1,ch=nc();  
	for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';  
	if (ch=='.'){  
		double tmp=1; ch=nc();  
		for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');  
	}  
	if (sign)x=-x;  
}  
inline void read(char *s){  
	char ch=nc();  
	for (;blank(ch);ch=nc());  
	if (IOerror)return;  
	for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;  
	*s=0;  
}  
inline void read(char &c){  
	for (c=nc();blank(c);c=nc());  
	if (IOerror){c=-1;return;}  
} 
#undef OUT_SIZE  
#undef BUF_SIZE  
}; using namespace fastIO;
/*****************************************************/

typedef set<int> Set;
map<Set,int>IDcache;	//集合映射为id；
vector<Set>Setcache;	//根据id取集合

int ID(Set x){			//查找集合x的ID，找不到就分配新id
	if(IDcache.count(x))return IDcache[x];
	Setcache.pb(x);
	return IDcache[x]=Setcache.size()-1;
}

#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())	//用于set_union系函数的首迭代器

int main(){
	stack<int>s;
	int n;cin>>n;
	while(n--){
		int m;cin>>m;
		for(int i=0;i<m;i++){
			string op;
			cin>>op;
			if(op[0]=='P')s.push(ID(Set()));
			else if(op[0]=='D')s.push(s.top());
			else{
				Set x1=Setcache[s.top()];s.pop();
				Set x2=Setcache[s.top()];s.pop();
				Set x;
				if(op[0]=='U')set_union(ALL(x1),ALL(x2),INS(x));
				if(op[0]=='I')set_intersection(ALL(x1),ALL(x2),INS(x));
				if(op[0]=='A'){x=x2;x.insert(ID(x1));}
				s.push(ID(x));
			}
			cout<<Setcache[s.top()].size()<<endl;
		}
		cout<<"***"<<endl;
	}
}