The SetStack Computer(UVa12096&&POJ3121) (集合栈)

The SetStack Computer

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1095 Accepted: 219

Description

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).

Sample Input

29PUSHDUPADDPUSHADDDUPADDDUPUNION5PUSHPUSHADDPUSHINTERSECT

Sample Output

001011222***00100***

   这是刘汝佳的算法入门经典上面的题。看到这道题的代码，我发现自己缺失很多知识。set，map，vector基本上是空白。

虽说是STL类型的题，但是还是要掌握的。任重而道远，立足当下。当意识到晚的时候其实还并不晚。

    分析：本题的集合并不是简单的整数集合或者是字符串集合，而是集合的集合。为了方便起见，此处为每个不同的集合

分配了唯一的ID，则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STL的set<int> 来表示，而整个栈则是一个

stack<int> 。

    对任意的集合s(Set 类型)，IDcache[s]就是它的ID，而Setcache[IDcache[s]]就是s本身!!!

代码：在Poj3121不知道为什么一直TLE

#include<iostream>#include<string>#include<vector>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;#define ALL(x)  x.begin(),x.end()#define INS(x)  inserter(x,x.begin())typedef set<int> Set;map<Set,int> IDcache; //把集合映射成ID vector<Set> Setcache; //根据ID取集合 //查找给定集合x的ID。如果查不到，分配一个新ID int ID(Set x){if(IDcache.count(x)) return IDcache[x];Setcache.push_back(x); //添加新集合 return IDcache[x]=Setcache.size()-1;}int main(){int n,test;cin>>test;while(test--){stack<int> s;cin>>n;for(int i=0;i<n;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;} }