1063 Set Similarity (25 分)
Given two sets of integers, the similarity of the sets is defined to be/, whereNcis the number of distinct common numbers shared by the two sets, andNtis 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 integerN(≤) which is the total number of sets. ThenNlines follow, each gives a set with a positiveM(≤) and followed byMintegers in the range [0]. After the input of sets, a positive integerK(≤) is given, followed byKlines of queries. Each query gives a pair of set numbers (the sets are numbered from 1 toN). 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:挺麻煩的一個題,充分利用了vector和set
#include<bits/stdc++.h> usingnamespace 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; }