1063 Set Similarity (25 分)
2021/6/28 23:20:33
本文主要是介绍1063 Set Similarity (25 分),对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
Given two sets of integers, the similarity of the sets is defined to be /, where Nc is the number of distinct common numbers shared by the two sets, and Nt 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 (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by M integers in the range [0]. After the input of sets, a positive integer K (≤) 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%思路:挺麻烦的一个题,充分利用了vector和set
#include<bits/stdc++.h> using namespace std; const int maxn=1010; #define eps 1e-8 #define inf 0x3fffffff int main(){ vector<set<int> > v; int n,k,m,t; cin>>n; v.resize(n); for(int i=0;i<n;i++){ cin>>m; for(int j=0;j<m;j++){ cin>>k; v[i].insert(k); } } cin>>t; int a,b; for(int i=0;i<t;i++){ cin>>a>>b; int ct=0,nt=v[a-1].size(); for(set<int>::iterator it=v[b-1].begin();it!=v[b-1].end();it++){ if(v[a-1].find(*it)!=v[a-1].end()){ ct++; } else{ nt++; } } printf("%.1f%\n",1.0*ct/nt*100); } return 0; }
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