C++PTAA1063 Set

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:

核心思想

这个stl没有python那样方便，将数据都一一录入进去，创造一个比较函数得出答案。

核心词汇

N_c is the number of distinct number shared by the two sets两个集合中相同的元素
N_t 两个集合中不同的数

完整代码

#include<cstdio>
#include<set>
using namespace std;
const int N = 51;
set<int> st[N];
void compare(int x,int y){
    int totalNum = st[y].size(),sameNum = 0;
    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]中能找该元素
        else totalNum++;//在st[y]中不能找到该元素
    }
    printf("%.1f%\n",sameNum*100.0/totalNum);//输出比率
}
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); //将元素v加入集合st[i]中
        }
    }

    scanf("%d",&q);
    for(int i =0;i<q;i++){
        scanf("%d%d",&st1,&st2);
        compare(st1,st2);//比较两个集合.
    }
    return 0;
}

测试效果