PAT 1063. Set Similarity (25) 寻找交集。未解之谜。

本文链接：https://blog.youkuaiyun.com/yxpandjay/article/details/77780665

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Nc为两个集合中共同拥有的不同整数的数量，Nt为两个集合中不同整数的总数，集合的相似度则为Nc/Nt*100%。文章提供了输入输出规范及一个具体的C++实现示例。

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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 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 (<=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%


不懂为啥把那两行注释掉就过了。感觉应该没影响的。。。
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<math.h>
#include<stack>
#include<vector>
using namespace std;

set<int> node[52];


void judge(int x,int y)
{
    set<int>::iterator it1=node[x].begin();
    set<int>::iterator it2=node[y].begin();
    int last=-1;
    int sum1=node[x].size()+node[y].size();
    int sum2=0;
    while(it1!=node[x].end()&&it2!=node[y].end())
      {
          if(*it1==*it2)
          {
              sum2++;
              //last=-1;
              it1++;
              it2++;
          }
          else if(*it1<*it2)
          {
           //  if(*it1==last) sum2++;    这两行一定要注释掉！！
             // last=*it1;
              it1++;
          }
          else { 
           //  if(*it2==last) sum2++;    这两行一定要注释掉！！
             // last=*it2;
              it2++;
                }
     }

  float tmp=(float)sum2/(float)(sum1-sum2);
             int zh=int(tmp*1000+0.5);
             int xx,yy,zz;
             xx=zh/100;
             yy=(zh%100)/10;
             zz=zh%10;
            if(xx!=0) cout<<xx;
            cout<<yy<<'.'<<zz<<'%'<<endl;

}


int main()
{
     int n,k;
     cin>>n;
    for(int i=1;i<=n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        for(int j=0;j<tmp;j++)
        {
            int tmmp;
            scanf("%d",&tmmp);
            node[i].insert(tmmp);
        }
    }
    cin>>k;
    for(int i=0;i<k;i++)
    {
        int a1,b1;
        scanf("%d %d",&a1,&b1);
        judge(a1,b1);
    }


     return 0;
}