1063. Set Similarity (25)

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最新推荐文章于 2023-05-18 23:02:00 发布

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分类专栏： C/C++ OJ pat甲级文章标签： pat甲级 1063

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本文介绍了一个计算两个整数集合相似度的问题，通过定义相似度为两集合共有不同整数的数量除以所有不同整数数量的比例，并提供了一种利用C++ set进行高效计算的方法。

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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 Nc/Nt*100%, where Nc is the number of distinct common numbers shared by the two sets, and Nt 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:
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%

思路：

用set去除重复的元素，同时set有自动升序的功能，因此可用用o(n)的方法实现查找相同的元素个数
也可用一个集合中的元素一个一个到另一个集合总去find，但是复杂度稍高，也可以通过

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <set>

using namespace std;

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    double rate[51][51];
    set<int> Data[51];
    set<int>::iterator s1, s2;
    int N, M, tmp;

    scanf("%d", &N);
    for (int i = 1; i <= N; ++i)
    {
        scanf("%d", &M);
        for (int k = 1; k <= M; ++k)
        {
            scanf("%d", &tmp);
            Data[i].insert(tmp);
        }
    }


    for (int i = 1, simil; i <= N; ++i)
    {
        for (int k = 1; k <= i; ++k)
        {
            if (k == i)continue;
            simil = 0; s1 = Data[i].begin(); s2 = Data[k].begin();
            while (s1 != Data[i].end() && s2 != Data[k].end())
            {
                if (*s1 == *s2)
                {
                    ++simil;
                    ++s1; ++s2;
                }
                else if (*s1 < *s2)
                    ++s1;
                else
                    ++s2;
            }

            rate[i][k] = (double)simil * 100 / (Data[i].size() + Data[k].size() - simil);
        }
    }


    scanf("%d", &N);
    while (N--)
    {
        int i, j;
        scanf("%d %d", &i, &j);
        if (i < j)
            swap(i, j);
        printf("%.1f%%\n", rate[i][j]);
    }

    return 0;
}