PAT甲级练习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%

一开始想用合并set后计算大小与原先两个大小的关系来计算，但最后一个超时了，改用遍历和find的方法来做，没想到这样反而耗时要少一点。。

#include <iostream>  #include <cstdio>  #include <algorithm>  #include <vector>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <string>  #include <string.h> using namespace std;const int MAX = 1e5+10;set<int> s[51];//void query(int a, int b){//int a1 = s[a].size(), b1 = s[b].size();//int c1 = a1 + b1;//set<int> x(s[a]);//x.insert(s[b].begin(), s[b].end());//int c = x.size();//printf("%.1lf%%\n", (c1-c)*1.0/c*100);//}int main() {int n, m, d, k, k1, k2;char c[10];scanf("%d", &n);for(int i=1; i<=n; i++){scanf("%d", &m);for(int j=0; j<m; j++){scanf("%d", &d);s[i].insert(d);}}scanf("%d", &k);for(int i=1; i<=k; i++){scanf("%d %d", &k1, &k2);int ns=0, nt=s[k2].size();for(auto it=s[k1].begin(); it!=s[k1].end(); it++){if(s[k2].find(*it)==s[k2].end())nt++;elsens++;}double ans = ns * 1.0 / nt *100;printf("%.1f%%\n", ans);}scanf("%d",&n);    return 0;}