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%

題目大意：給定兩個整數集合，它們的相似度定義為：Nc/Nt*100%。其中Nc是兩個集合都有的不相等整數的個數，Nt是兩個集合一共有的不相等整數的個數。你的任務就是計算任意一對給定集合的相似度。

分析：因為給出的集合裡面含有重複的元素，而計算nc和nt不需要考慮兩個集合裡面是否分別有重複的元素，所以可以直接使用set儲存每一個集合，然後把set放進一個數組裡面儲存。當需要計算集合a和集合b的相似度nc和nt的時候，遍歷集合a中的每一個元素，尋找集合b中是否有該元素，如果有，說明是兩個人公共的集合元素，則nc++，否則nt++（nt的初值為b集合裡面本有的元素）

PS: 注意使用set的查詢find函式

#include<cstdio>
#include<set>
#include<vector>
using namespace std;

int main(){
	int n,m,k,num,nc,nt;
	scanf("%d",&n);
	vector<set<int>> s(n+1);
	for(int i=1;i<=n;i++){
		scanf("%d",&m);
		for(int j=1;j<=m;j++){
			scanf("%d",&num);
			s[i].insert(num);
		}
	}
	scanf("%d",&k);
	int a,b;
	for(int i=1;i<=k;i++){
        scanf("%d%d",&a,&b);
        int nc = 0,nt = s[b].size();
        for(auto ita = s[a].begin();ita!=s[a].end();ita++)
        {
            if(s[b].find(*ita) != s[b].end())//ͨ¹ýfind²éÕÒ 
            	nc++;
            else
            	nt++;
        }
        printf("%.1f%\n",double(nc)/nt * 100);
	}
	return 0;
}