1063 Set Similarity (25 分)

1063 Set Similarity (25 分)

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% #include <bits/stdc++.h> using namespace std; const int N=51; set<int>st[N]; void compare(int x,int y) { int totalNum=st[y].size(); int sameNum=0; for(set<int>::iterator it=st[x].begin(); it!=st[x].end(); it++) { if(st[y].find(*it)!=st[y].end())sameNum++; else totalNum++; } //printf("%d %d\n",sameNum,totalNum); printf("%.1lf%%\n",sameNum*100.0/totalNum); } int main() { int n,k,q,v,st1,st2; scanf("%d",&n); for(int i=1; i<=n; i++) { scanf("%d",&k); for(int j=0; j<k; j++) { scanf("%d",&v); st[i].insert(v); } } scanf("%d",&q); for(int i=0; i<q; i++) { scanf("%d%d",&st1,&st2); compare(st1,st2); } return 0; }