1063. Set Similarity

1063. Set Similarity (25)

时间限制

300 ms

内存限制

65536 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%

输入排好序，做类似于两个有序数组合并的操作，多了一个当前值和上一个值是否相等的判断条件。

#include <iostream>#include <vector>#include <algorithm>using namespace std;int main(){int n,k;cin>>n;vector<vector<int>> num(n);for(int i=0;i<n;++i){int m;cin>>m;for(int j=0;j<m;++j){int t;cin>>t;num[i].push_back(t);}sort(num[i].begin(),num[i].end());}cin>>k;for(int i=0;i<k;++i){int p1,p2;cin>>p1>>p2;p1--;p2--;int q1=0,q2=0;double nc=0,nt=0;int tn=-1;while(q1<num[p1].size()&&q2<num[p2].size()){if(num[p1][q1]<num[p2][q2]){if(num[p1][q1]!=tn){nt++;tn=num[p1][q1];}q1++;}else if(num[p1][q1]>num[p2][q2]){if(num[p2][q2]!=tn){nt++;tn=num[p2][q2];}q2++;}else {if(tn!=num[p1][q1]){nt++;nc++;tn=num[p1][q1];}q1++;q2++;}}while(q1<num[p1].size()){if(num[p1][q1]!=tn){tn=num[p1][q1];nt++;}q1++;}while(q2<num[p2].size()){if(num[p2][q2]!=tn){tn=num[p2][q2];nt++;}q2++;}printf("%.1lf",nc/nt*100);cout<<"%"<<endl;}return 0;}