Pat(Advanced Level)Practice--1063(Set Similarity)

Pat1063代码

题目描述：

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:

33 99 87 1014 87 101 5 877 99 101 18 5 135 18 9921 21 3

Sample Output:

50.0%33.3%

利用map和set的性质进行筛选，结果最后一个case超时。

代码：

#include<cstdio>#include<map>#include<set>#define N 55using namespace std;set<int> s[N];int main(int argc,char *argv[]){int n;int i,j;scanf("%d",&n);for(i=1;i<=n;i++){int m;int temp;scanf("%d",&m);for(j=1;j<=m;j++){scanf("%d",&temp);s[i].insert(temp);}}int k;scanf("%d",&k);while(k--){map<int,int> m;set<int>::iterator p;int x,y;scanf("%d%d",&x,&y);for(p=s[x].begin();p!=s[x].end();p++){int index=*p;m[index]++;}for(p=s[y].begin();p!=s[y].end();p++){int index=*p;m[index]++;}int count=0;map<int,int>::iterator it;for(it=m.begin();it!=m.end();it++)if(it->second>1)count++;printf("%.1f%%\n",100*count*1.0/m.size());}return 0;}

AC代码：

利用函数求两个集合的交集，然后就可以求出集合的相似性了；

#include<cstdio>#include<set>#include<iterator>#include<algorithm>#define N 55using namespace std;set<int> s[N];int main(int argc,char *argv[]){int n;int i,j;scanf("%d",&n);for(i=1;i<=n;i++){int m;int temp;scanf("%d",&m);for(j=1;j<=m;j++){scanf("%d",&temp);s[i].insert(temp);}}int k;scanf("%d",&k);while(k--){int x,y;int ret,total;set<int> ans;scanf("%d%d",&x,&y);set_intersection(s[x].begin(),s[x].end(),s[y].begin(),s[y].end(),inserter(ans,ans.begin()));//求集合交集ret=ans.size();//交集元素个数total=s[x].size()+s[y].size()-ret;//不重复元素总个数printf("%.1f%%\n",100*ret*1.0/total);}return 0;}

最后一个case跑了大概130ms;

AC代码：

写一个函数求集合的并集，其实这和上面的方法是一样的，只不过一个是交集，一个是并集；

#include<cstdio>#include<set>#include<iterator>#include<algorithm>#define N 55using namespace std;set<int> s[N];int GetUnion(int x,int y)//求并集元素的个数，其实algorithm里面有set_union{                        //这个函数，在这里我们可以简单的实现一下int ret=0;set<int>::iterator it;for(it=s[x].begin();it!=s[x].end();it++){if(s[y].find(*it)==s[y].end())ret++;}ret+=s[y].size();return ret;}int main(int argc,char *argv[]){int n;int i,j;scanf("%d",&n);for(i=1;i<=n;i++){int m;int temp;scanf("%d",&m);for(j=1;j<=m;j++){scanf("%d",&temp);s[i].insert(temp);}}int k;scanf("%d",&k);while(k--){int x,y;int ret,total;scanf("%d%d",&x,&y);total=GetUnion(x,y);ret=s[x].size()+s[y].size()-total;printf("%.1f%%\n",100*ret*1.0/total);}return 0;}

最后一个case跑了大概170ms;