算法竞赛--set的使用

1.set的定义

set<typename> name;

2.vector的遍历

set<int> s;
for(set<int>::iterator it=s[num].begin();it!=s[num].end();it++)
	printf(" %d",*it);

3.常用函数

1.在set后面添加一个元素 insert()

set<int> s;
for(int i=0;i<10;i++)
	s.insert(i);

2.求set的长度

set<int> s;
s.size();

3.清空set中所有元素

set<int> s;
s.clear();

4.查找元素(返回set中对应值为value的迭代器)

set<int> s;
s.find();

4.删除单个元素

set<int> s;
s.erase(100);	//删除值为100的元素

7.删除一个区间内所有元素 注意左闭右开

set<int> s;
s.erase(s.begin(),s.end());	//删除所有元素

Set Similarity

Given two sets of integers, the similarity of the sets is defined to be Nc/Nt×100%, where ​c 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.

输入格式:

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.

输出格式:

For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3

输出样例:

50.0%
33.3%

思路:

用set解题 求两个集合的交集和并集

我的代码如下:

#include<bits/stdc++.h>
using namespace std;
const int N=55;
set<int>s[N];
void compare(int a,int b)
{
	int tn=s[a].size(),sn=0;//分别记录并集和交集元素的个数
	for(set<int>::iterator it=s[b].begin();it!=s[b].end();it++)
	{
		if(s[a].find(*it)!=s[a].end())
			sn++;
		else
			tn++;
	}
	printf("%.1f%%",sn*100.0/tn);//%%输出 %号
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int m,x;
		scanf("%d",&m);
		for(int j=0;j<m;j++)
		{
			scanf("%d",&x);
			s[i].insert(x);
		}	
	}
	int k,x,y;
	scanf("%d",&k);
	for(int i=0;i<k;i++)
	{
		scanf("%d%d",&x,&y);
		compare(x,y);
		if(i!=k-1)
			printf("\n");
	}
	return 0;
}