1063. Set Similarity (25)

最新推荐文章于 2018-03-07 20:48:34 发布

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本文介绍了一种计算两个整数集合相似度的方法，并通过C++代码实现。利用set_intersection函数求交集，计算交集与并集元素的比例，得出集合相似度。

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Given two sets of integers, the similarity of the sets is defined to be N_c/N_t*100%, where N_c is the number of distinct common numbers shared by the two sets, and N_t is the total number of distinct numbers in the two sets. Your job is to calculate the similarity of any given pair of sets.

Input Specification:

Each input file contains one test case. Each case first gives a positive integer N (<=50) which is the total number of sets. Then N lines follow, each gives a set with a positive M (<=10⁴) and followed by M integers in the range [0, 10⁹]. After the input of sets, a positive integer K (<=2000) is given, followed by K lines of queries. Each query gives a pair of set numbers (the sets are numbered from 1 to N). All the numbers in a line are separated by a space.

Output Specification:

For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

Sample Input:

3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3

Sample Output:

50.0%
33.3%

考察集合的应用，关键要知道用函数的使用

集合介绍：set集合容器实现了红黑树（Red-Black Tree）的平衡二叉检索树的数据结构，在插入数据时会自动调整二叉树的排列确保每个子树根节点的键值都大于左子树所有节点的键值，小于右子树所有节点的键值！

参考AC代码：

#include <iostream>
#include <algorithm>		//set_intersection 函数
#include <iomanip>
#include <iterator>		//inserter函数
#include <set>
using namespace std;

int main()
{
    int n,m,in;
    cin>>n;
    set<int> arr[n];
    int i,j;
    for(i=0;i<n;i++)
    {
        cin>>m;
		for(j=0;j<m;j++)
		{
			cin>>in;
			arr[i].insert(in);
		}
   }
   set<int> destination;
   int k,a,b,interSize,totalSize;
   cin>>k;
   for(i=0;i<k;i++)
   {
		cin>>a>>b;
		set_intersection(arr[a-1].begin(),arr[a-1].end(),arr[b-1].begin(),arr[b-1].end(),inserter(destination,destination.begin()));		//求集合交集
		interSize = destination.size();
		totalSize = arr[a-1].size()+arr[b-1].size()-interSize;
		destination.clear();

		cout<<fixed<<setprecision(1)<<((double)interSize/totalSize)*100.0<<"%"<<endl;
   }
    return 0;
}

主要对set_intersection函数作下简单的介绍：

这个函数是求集合的交集的

template <class InputIterator1, class InputIterator2, class OutputIterator>
  OutputIterator set_intersection (InputIterator1 first1, InputIterator1 last1,
                                   InputIterator2 first2, InputIterator2 last2,
                                   OutputIterator result)
{
  while (first1!=last1 && first2!=last2)
  {
    if (*first1<*first2) ++first1;
    else if (*first2<*first1) ++first2;
    else {
      *result = *first1;
      ++result; ++first1; ++first2;
    }
  }
  return result;
}

下面来一个简单的例子作为理解：

// set_intersection example
#include <iostream>     // std::cout
#include <algorithm>    // std::set_intersection, std::sort
#include <vector>       // std::vector

int main () {
  int first[] = {5,10,15,20,25};
  int second[] = {50,40,30,20,10};
  std::vector<int> v(10);                      // 0  0  0  0  0  0  0  0  0  0
  std::vector<int>::iterator it;

  std::sort (first,first+5);     //  5 10 15 20 25
  std::sort (second,second+5);   // 10 20 30 40 50

  it=std::set_intersection (first, first+5, second, second+5, v.begin());
                                               // 10 20 0  0  0  0  0  0  0  0
  v.resize(it-v.begin());                      // 10 20

  std::cout << "The intersection has " << (v.size()) << " elements:\n";
  for (it=v.begin(); it!=v.end(); ++it)
    std::cout << ' ' << *it;
  std::cout << '\n';

  return 0;
}

Output：

The intersection has 2 elements:
 10 20