1063 Set Similarity (25 分)【STL中vector與set的使用】
1063 Set Similarity (25 分)
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%
題意:公共元素個數除以2個set不同元素的總個數,如例子:set1和set2中2個公共元素,總共4個不相同的元素
解題思路:這個一看就是不要重複的,那肯定使用set比較方便,但是有多個set,所以用vector來儲存這些set,然後我們直接從b找a的元素,找得到就是count1加1即公共元素,找不到就count2加1,那這2個set中總共count2+b的個數,應該get得到吧,具體實現在下面
#include<bits/stdc++.h>
using namespace std;
vector<set<int> >v;
int main(void)
{
int n;
scanf("%d",&n);
v.resize(n+1);
for(int i=1;i<=n;i++)
{
int m;
scanf("%d",&m);
for(int j=0;j<m;j++)
{
int a;
scanf("%d",&a);
v[i].insert(a);
}
}
int k,a,b;
scanf("%d",&k);
while(k--)
{
scanf("%d %d",&a,&b);
int len1=v[a].size(),len2=v[b].size(),count1=0,count2=0;
// cout<<len1<<" "<<len2<<endl;
for(auto i=v[a].begin();i!=v[a].end();i++)
{
// cout<<*i<<endl;
if(v[b].count(*i)!=0) count1++;
else count2++;
}
// cout<<count1<<" "<<count2<<" "<<count2+len2<<endl;
printf("%.1lf%\n",count1*1.0/(count2+len2)*100);
}
return 0;
}