EOJ 1641/UVa The SetStack Computer
Background from Wikipedia: “Set theory is a branch of mathematics created principally by the German mathematician Georg Cantor at the end of the 19th century. Initially controversial, set theory has come to play the role of a foundational theory in modern mathematics, in the sense of a theory invoked to justify assumptions made inmathematics concerning the existence of mathematical objects (such as numbers or functions) and their properties. Formal versions of set theory also have a foundational role to play as specifying a theoretical ideal of mathematical rigor in proofs.”
Given this importance of sets, being the basis of mathematics, a set of eccentric theorist set off to construct a supercomputer operating on sets instead of numbers. The initial Set-Stack Alpha is under construction, and they need you to simulate it in order to verify the operation of the prototype.
The computer operates on a single stack of sets, which is initially empty. After each operation, the cardinality of the topmost set on the stack is output. The cardinality of a set S is denoted |S| and is the number of elements in S. The instruction set of the SetStack Alpha is PUSH, DUP, UNION, INTERSECT, and ADD.
PUSH will push the empty set {} on the stack.
DUP will duplicate the topmost set (pop the stack, and then push that set on the stack twice).
UNION will pop the stack twice and then push the union of the two sets on the stack.
INTERSECT will pop the stack twice and then push the intersection of the two sets on the stack.
ADD will pop the stack twice, add the first set to the second one, and then push the resulting set on the stack.
For illustration purposes, assume that the topmost element of the stack is
A = {{}, {{}}}
and that the next one is
B = {{}, {{{}}}}.
For these sets, we have |A| = 2 and |B| = 2. Then:
◎ UNION would result in the set { {}, {{}}, {{{}}} }. The output is 3.
◎ INTERSECT would result in the set { {} }. The output is 1.
◎ ADD would result in the set { {}, {{{}}}, {{},{{}}} }. The output is 3.
Input
An integer 0 ≤ T ≤ 5 on the first line gives the cardinality of the set of test cases. The first line of each test case contains the number of operations 0 ≤ N ≤ 2 000. Then follow N lines each containing one of the five commands. It is guaranteed that the SetStack computer can execute all the commands in the sequence without ever popping an empty stack.
Output
For each operation specified in the input, there will be one line of output consisting of a single integer. This integer is the cardinality of the topmost element of the stack after the corresponding command has executed. After each test case there will be a line with *** (three asterisks).
Examples
Input
2
9
PUSH
DUP
ADD
PUSH
ADD
DUP
ADD
DUP
UNION
5
PUSH
PUSH
ADD
PUSH
INTERSECT
Output
0
0
1
0
1
1
2
2
2
***
0
0
1
0
0
***
1 #include <iostream>
2 #include <map>
3 #include <set>
4 #include <algorithm>
5 #include <stack>
6 #define ALL(x) x.begin(),x.end()
7 #define INS(x) inserter(x,x.begin())
8 using namespace std;
9 typedef set<int> Set;
10 map<Set,int> IDcache;
11 vector<Set> Setcache;
12 int ID(Set x){
13 if(IDcache.count(x)) return IDcache[x];
14 Setcache.push_back(x);
15 return IDcache[x]=Setcache.size()-1;
16 }
17 int main()
18 {
19 int n,ord;cin>>n;
20 stack<int> s;
21 while(n--){
22 cin>>ord;
23 while(ord--){
24 string op;
25 cin>>op;
26 if(op[0]=='P') s.push(ID(Set()));
27 else if(op[0]=='D') s.push(s.top());
28 else{
29 Set x1=Setcache[s.top()];s.pop();
30 Set x2=Setcache[s.top()];s.pop();
31 Set x;
32 switch(op[0]){
33 case 'U': set_union(ALL(x1),ALL(x2),INS(x));break;
34 case 'I': set_intersection(ALL(x1),ALL(x2),INS(x));break;
35 case 'A': x=x2;x.insert(ID(x1));break;
36 }
37 s.push(ID(x));
38 }
39 cout << Setcache[s.top()].size()<<endl;
40 }
41 cout<<"***"<<endl;
42
43 }
44
45 return 0;
46 }
本题的集合并不是简单的整数集合或字符串集合，而是集合的集合。map为每个不同的集合分配一个唯一的ID，每个key的value是key这个集合的ID，每个集合都可以表示成所包含元素的ID集合，这样就可以用set<int>表示，而整个栈则是一个stack<int>，每一次将集合的ID推入。vec数组方便根据ID取集合。
26行的ID(Set())应该是空集的ID，为0。
举个例子，
第一次PUSH，ID()为空集分配ID——0，并保存入map中，栈推入空集的ID：0；
第二次DUP，将栈顶的集合ID：0再推入栈；
第三次ADD，出栈两个元素，都是空集，ID均为0，将第一个集合加入第二个集合里，即是将第一个集合ID插入到第二个集合中，并给新集合：{0}分配ID——1；并将新集合ID推入栈。则栈顶集合ID：1，集合内元素{0}，元素个数:1。
第一次 map:{} 0 vector [0]: 空 stack:0
第二次 map:{} 0 vector [0]: 空 stack:0 0
第三次 map:{} 0，{0}，1 vector [0]: 空，[1]:{0} stack:0 0 1
感觉说的还不是很清楚，多读代码理解吧！
