1063. Set Similarity (25)

题目：

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%

注意：

1、直接使用set容器，把重复的剔除掉。

2、然后merge归并，计出重复的个数和总个数。

代码：

#include<iostream>
#include<vector>
#include<set>
using namespace std;

int main()
{
	vector<set<int>>allset;
	int n;
	scanf("%d",&n);
	allset.resize(n);
	int max;
	for(int i=0;i<n;++i)
	{
		int k;
		scanf("%d",&k);
		for(int j=0;j<k;++j)
		{
			int t;
			scanf("%d",&t);
			allset[i].insert(t);
		}
	}
	int m;
	scanf("%d",&m);
	for(int i=0;i<m;++i)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		--a; --b;//the i-th set is in set[i-1]
		float nt=0,nc=0;
		set<int>::iterator a_iter,b_iter;
		a_iter=allset[a].begin();
		b_iter=allset[b].begin();
		int cur=-1;
		for(int j=0;j<allset[a].size()+allset[b].size();++j)
		{//merge
			int next;
			if(a_iter!=allset[a].end() && b_iter!=allset[b].end())
			{
				if(*a_iter<*b_iter)
					next=*a_iter++;
				else
					next=*b_iter++;
			}
			else if(a_iter==allset[a].end())
				next=*b_iter++;
			else if(b_iter==allset[b].end())
				next=*a_iter++;
			if(next!=cur)
				++nt;
			else
				++nc;
			cur=next;
		}
		printf("%.1f%%\n",100*nc/nt);
	}
	return 0;
}