【PAT】【Advanced Level】1063. Set Similarity (25)

1063. Set Similarity (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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%

原题链接：

https://www.patest.cn/contests/pat-a-practise/1063

https://www.nowcoder.com/pat/5/problem/4114

思路：

先使每一个数列有序

然后采取归并的方法进行统计

CODE：

#include<iostream>#include<vector>#include<cstring>#include<string>#include<algorithm>#include<cstdio>using namespace std;vector<int> v[51];bool cmp(int a,int b){return a<b;}int main(){int n;//cin>>n;scanf("%d",&n);for (int i=0;i<n;i++){int m;//cin>>m;scanf("%d",&m);for (int j=0;j<m;j++){int t;//cin>>t;scanf("%d",&t);v[i].push_back(t);}sort(v[i].begin(),v[i].end(),cmp);}int k;//cin>>k;scanf("%d",&k);for (int i=0;i<k;i++){int a,b;//cin>>a>>b;scanf("%d%d",&a,&b);a--;b--;int t1=0;int t2=0;int p1=0;int p2=0;while (p1<v[a].size() && p2<v[b].size()){//cout<<v[a][p1]<<" "<<v[b][p2]<<endl;if (p1>0&&v[a][p1]==v[a][p1-1]){p1++;continue;}if (p2>0&&v[b][p2]==v[b][p2-1]){p2++;continue;}if (v[a][p1]<v[b][p2]){t2++;p1++;}elseif (v[a][p1]>v[b][p2]){t2++;p2++;}elseif (v[a][p1]==v[b][p2]){t1++;t2++;p1++;p2++;}}if (p1<v[a].size()){for (int j=p1;j<v[a].size();j++){if (j==0 || v[a][j]!=v[a][j-1]) t2++;}}if (p2<v[b].size()){for (int j=p2;j<v[b].size();j++){if (j==0 || v[b][j]!=v[b][j-1]) t2++;}}double ss=(t1*1.0)/(t2*1.0);ss=ss*100;printf("%.1lf%c\n",ss,'%');}return 0;}