【PAT】A1063 Set Similarity

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 Nlines 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%

解题思路：

题目大意：给出n个序列和其中的数，要求给出指定两个集合的(相同个数)/(不同个数)的比率
使用STL中的set容器，因为set容器自动去掉重复的数并且按升序排序集合里的数。代码分为2段：首先按照要求输入数据，即输入“要输入的集合数n”和“n个集合，每个集合要输入的数m，以及输入m个数”，用set的insert()函数可以过滤掉重复的数据；下一步是整理数据，例如统一确定st[set2]是被查找的集合，循环st[set1]，如果找到st[set2].find(*it)!=st[set2].end()，意思是非空，也就是找到了，就相同数加一，反之就是st[set1]中有的数在st[set2]中是没有的，那么把st[set2].size()++，每一次输出即可。
总结：在st[set2].find(*it)!=st[set2].end()，这里需要思路清晰，大脑不能犯糊涂，对照1查2,2中查不到就++，不是对照1查1

代码：

#include <stdio.h>
#include <set>
using namespace std;
set<int> st[55];int main(){int n,m,x,set1,set2,inquiry,same,count=0;scanf("%d",&n); //集合的数量n//输入数据 for(int i=1;i<=n;i++){scanf("%d",&m);  //集合中的数量mfor(int j=0;j<m;j++){scanf("%d",&x);  //集合中的每一个数，赋值为xst[i].insert(x);  //将数插入到第i个集合中 } }//整理数据scanf("%d",&inquiry); //需要查询的次数for(int i=0;i<inquiry;i++){same=0;scanf("%d%d",&set1,&set2);  //比较的集合编号int count=st[set2].size();for(set<int>::iterator it=st[set1].begin();it!=st[set1].end();it++){//从第一个集合中的每一个数，看是否有数和st[set2]中的数一致，一致就same++,不一致则count++if(st[set2].find(*it)!=st[set2].end()){//一开始弄错的地方same++;//printf("%d\n",same);}else{count++;//printf("c%d\n",count);}}printf("%.1lf%%\n",(double)(same*100.0/count));}return 0;
}