PAT练习笔记——6.2 set的常见用法详解

2019年9月PAT - 练习笔记——6.2

以下页码标注的是阅读器中实际页码，而不是书本身自印的页码。

第6章 C++标准模板库（STL）介绍

6.2 set的常见用法详解

注意

1. std::set和std::multiset为有序序列，而hash_set以及hash_multiset为无序序列。

2. 并交差：https://blog.csdn.net/u013095333/article/details/89322501

   1. set_union(A.begin(),A.end(),B.begin(),B.end(),inserter( C1 , C1.begin() ) );
   2. set_intersection(A.begin(),A.end(),B.begin(),B.end(),inserter( C2 , C2.begin() ));
   3. set_difference( A.begin(), A.end(),B.begin(), B.end(),inserter( C3, C3.begin() ) );

   注意！以上函数要求两个集合必须是有序的

目录

1. A1063 Set Similarity

---

1. A1063 Set Similarity

   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%
   

   1. 我的

      #include 
      #include 
      #include 
      
      using namespace std;
      
      const int INF = 0x7fffffff;
      
      int main(void)
      {
      	int n = 0;
      	scanf("%d", &n);
      	
      	vector> sets(n + 1);
      	for (int i = 1;i <= n;++i) {
      		int m = 0;
      		scanf("%d", &m);
      		for (int j = 0;j < m;++j) {
      			int num = 0;
      			scanf("%d", &num);
      			sets[i].insert(num);
      		}
      		sets[i].insert(INF);
      	}
      	
      	int k = 0;
      	scanf("%d", &k);
      	for (int i = 0;i < k;++i) {
      		int set1 = 0, set2 = 0;
      		scanf("%d %d", &set1, &set2);
      		
      		int count1 = -1, count2 = -1;
      		set::iterator iter1 = sets[set1].begin(), iter2 = sets[set2].begin();
      		for (;iter1 != sets[set1].end() && iter2 != sets[set2].end();) {
      			int num1 = *iter1, num2 = *iter2;
      			if (num1 == num2) {
      				++count1;
      				++iter1;
      				++iter2;
      			}
      			else if (num1 < num2) ++iter1;
      			else ++iter2;
      			
      			++count2;
      		}
      		
      		float percent = (float)count1 / count2;
      		printf("%.1f%%\n", percent * 100);
      	} 
      	
      	return 0;
      } 
      

      用set_union和set_intersection的话，测试点4会超时

   2. 《算法笔记》P246