1063 Set Similarity
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%
题意： 给出N个集合，给出的集合中可能含有相同的值。然后要求M个查询，每个查询给出两个集合的编号X和Y，求集合X和集合Y的相同元素率，即两个集合的交集和并集（均需去重）的元素个数的比率参考代码：#include<cstdio>#include<set>using namespace std;const int N = 51;set<int> st[N]; //N个集合void compare(int x, int y) { //比较集合st[x]和集合st[y] int totalNum = st[y].size(), sameNum = 0; //不同数的个数，相同数个个数 //遍历集合st[x] for(set<int>::iterator it = st[x].begin(); it != st[x].end(); it++) { if(st[y].find(*it) != st[y].end()) sameNum++; //在st[y]中找到相同该元素 else totalNum++; //在st[y]中找不到相同元素 } printf("%.1f%%\n", 100 * (double)sameNum/totalNum); //输出比率}
int main(){ int n,k,q,v,st1,st2; scanf("%d", &n); //集合个数 for(int i = 1; i <= n; i++) { scanf("%d", &k); //集合i中的元素个数 for(int j = 0; j < k; j++) { scanf("%d", &v); //集合i中的元素v st[i].insert(v); //将元素v加入集合st[i]中 } } scanf("%d", &q); //q个查询 for(int i = 0; i < q; i++) { scanf("%d%d", &st1, &st2); //欲对比的集合编号 compare(st1, st2); //比较两个集合 } return 0;}
原文地址：https://www.cnblogs.com/mxj961116/p/10348228.html