1063 Set Similarity (25)

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%

这种方法有一个测试点不能通过，原因是建立map的开销比较大，并且再查询的时候，会反复的建立map，是一个很耗费时间的过程，应该在输出数据的时候就为每一组数据建立map，这样能避免之后重复建立map；
在编程中应该避免重复没有必要的操作

#include<iostream>
#include<vector>
#include<map>
#include<set>
using namespace std;
int main(){
  int n, m, i, j;
  vector<int> v[51];
  cin>>n;
  for(i=0; i<n; i++){
    cin>>m;
    for( j=0; j<m; j++){
      int temp;
      scanf("%d", &temp);
      v[i].push_back(temp);
    }
  }
  cin>>m;
  for(i=0; i<m; i++){
    int a, b, cnt=0, sum=0;
    cin>>a>>b;
    map<int, int> mmap;
    for(j=0; j<v[a-1].size(); j++)
        if(mmap.count(v[a-1][j])==0) mmap[v[a-1][j]] = 1;
    for(j=0; j<v[b-1].size(); j++){
        if(mmap.count(v[b-1][j])==0) mmap[v[b-1][j]] = 2;
        else if(mmap[v[b-1][j]]==1){
            cnt++;
            mmap[v[b-1][j]]=2;
        }
    }
    printf("%.1f%%
", cnt*100.0/mmap.size());
  }
  return 0;
}

修改：在输入数据的时候，就组织数据， 并用更简洁的set来建立数据结构

#include<iostream>
#include<set>
#include<vector>
using namespace std;
int main(){
  int n, m, i,j;
  cin>>n;
  vector<set<int>> v;
  for(i=0; i<n; i++){
    set<int> s;
    cin>>m;
    int temp;
    for(j=0; j<m; j++){
      cin>>temp;
      s.insert(temp);
    }
    v.push_back(s);
  }
  cin>>m;
  for(i=0; i<m; i++){
    int a, b;
    cin>>a>>b;
    int nc=v[a-1].size(), nt=0;
    for(auto it=v[b-1].begin(); it!=v[b-1].end(); it++){
      if(v[a-1].count(*it)!=0) nt++;
      else nc++;
    }
    printf("%.1f%%
", 100.0*nt/nc);
  }
  return 0;
}