PAT 1063. Set Similarity (25)

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最新推荐文章于 2022-05-05 19:26:29 发布

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本文介绍了一个简单的算法问题，即计算两个整数集合之间的相似度。输入包括多个集合及其元素，随后进行查询，输出每对集合的相似度百分比。文章提供了两种实现方式，并对比了它们的效率。

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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%

题目很简单，就是求两个集合的并的个数和两个集合交的个数。
写了两份代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;

int main()
{
  int n;
  while(scanf("%d",&n)!=EOF)
  {
    int group[52],sets[52][104];
    map<int,bool> allNum[52];
    for(int i = 0; i < n; i++)
    {
      int num = 0,x;
      map<int,bool> number;
      scanf("%d",&group[i]);
      for(int j = 0; j < group[i]; j++)
      {
        scanf("%d",&x);
        if(number[x] == false)
        {
          number[x] = true;
          sets[i][num++] = x;
          allNum[i][x] = true;
        }
      }
      group[i] = num;
    }
    int K;
    scanf("%d",&K);
    while(K--)
    {
      int a,b;
      int n1 = 0,n2 = 0;
      map<int,int> Bnumber;
      scanf("%d%d",&a,&b);
      a = a - 1;
      b = b - 1;
      n2 = group[a] + group[b];
      if(group[a] < group[b])
      {
        int temp = a;
        a = b;
        b = temp;
      }
      //可以替换下面几种
      map<int,bool>::iterator it;
      for(it=allNum[b].begin();it!=allNum[b].end();++it)
      {
        if(allNum[a].find(it->first) != allNum[a].end())
         n1++;
      }
       printf("%.1lf%%\n",(n1 * 1.0)/(n2-n1) * 100);
    }
  }
  return 0;
}

替换的地方换成一下代码，会超时，求助~

for(int i = 0; i < group[b]; i++)
 {
    if(allNum[a][sets[b][i]] == true)
         n1++;
 }

这种并不会超时，说明find函数比allNum[a][sets[b][i]]效率高吗？

for(int i = 0; i < group[b]; i++)
 {
    if(allNum[a].find(sets[b][i]) != allNum[a].end())
         n1++;
 }