【PAT1063】Set Similarity 求两集合的交集、并集

1063. Set Similarity (25)

时间限制

300 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

33 99 87 1014 87 101 5 877 99 101 18 5 135 18 9921 21 3

Sample Output:

50.0%33.3%

题意：求两个集合的交集/并集*100%

分析：使用set可以在insert时直接过滤掉重复的数值；使用set_intersection函数可以求两集合交集，set_union函数求并集。

代码：

#include <iostream>#include <fstream>#include <algorithm>//set_intersection 函数#include <iomanip>#include <iterator>//inserter函数#include <set>using namespace std;#include <stdio.h>//此代码使用前，需删除下面两行+后面的system("PAUSE")ifstream fin("in.txt");#define cin finint main(){   int n,m,in;   cin>>n;   set<int> *s = new set<int>[n];   int i,j;   for(i=0;i<n;i++)   {cin>>m;for(j=0;j<m;j++){cin>>in;s[i].insert(in);}   }   set<int> res;   int k,a,b,nc,nt;   cin>>k;   for(i=0;i<k;i++)   {cin>>a>>b;set_intersection(s[a-1].begin(),s[a-1].end(),s[b-1].begin(),s[b-1].end(),inserter(res,res.begin()));//求集合交集nc = res.size();nt = s[a-1].size()+s[b-1].size()-nc;res.clear();//set_union(s[a-1].begin(),s[a-1].end(),s[b-1].begin(),s[b-1].end(),inserter(res,res.begin()));//求集合并集，可通过数值计算得出//nt = res.size();//res.clear();cout<<setiosflags(ios::fixed)<<setprecision(1)<<(double)nc*100/nt<<"%"<<endl;   }    system("PAUSE");    return 0;}