PAT 1063 Set Similarity [set]

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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 (≤10​4​​) and followed by M integers in the range [0,10​9​​]. 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%

----------------------------------这是题目和解题的分割线----------------------------------

求交集/并集，所以要用set剔除重复元素。

#include<cstdio>
#include<set>using namespace std;int main()
{set<int> s[54];int n,x,i,num;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&x);while(x--){scanf("%d",&num); s[i].insert(num);}}int q,a,b;scanf("%d",&q);while(q--){scanf("%d%d",&a,&b);int count = 0;//set只能通过迭代器访问，不可以数组下标 for(set<int>::iterator it=s[a].begin();it!=s[a].end();it++)//没找到 !=s[b].end()（左闭右开end不存数据） if(s[b].find(*it)!=s[b].end()) count++;//并集元素=两个集合相加-重复元素 printf("%.1f%%\n",1.0*100*count/(s[a].size()+s[b].size()-count));}
} 

 

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