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%

总共N个集合；

每个集合的【初始】元素个数（有重复）   集合元素；【Ps:集合编号1~N，这里输入的时候要注意去除重复，我用set.inset(),会自动去重】

……

接下来K行

集合a 集合b 

……

输出K行，每对集合的    交集元素个数（相同原数个数）  占  并集元素个数（两个集合元素合在一起去重复以后的元素个数）  的百分比

下面两个代码，一个ac另一个最后一个测试点超时

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月13日 10:36答案正确251063C++ (g++ 4.7.2)1472744datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确118012/121答案正确13083/32答案正确11803/33答案正确64363/34答案正确14727444/4

#include <iostream> #include <vector>#include<set> using namespace std;void readlnsets(vector<set<int>>*sets, int N){  int M, integers,index;  for (index = 0; index < N; index++)  {    scanf("%d", &M);    while (M--)    {      scanf("%d", &integers);      (*sets)[index].insert(integers);/*当数已经在集合里面就不再放入了*/    }   }}int main(){  int N,K,a,b;  double Nt, Nc;   scanf("%d", &N);  vector<set<int>>sets(N);  readlnsets(&sets,N);  scanf("%d", &K);  while (K--)  {    scanf("%d%d", &a, &b);    a--;     b--;    Nc = 0.0;    for (set<int>::iterator iter = sets[a].begin(); iter != sets[a].end(); iter++)      if (sets[b].find((*iter))!= sets[b].end())Nc++;     Nt = sets[a].size() + sets[b].size()-Nc;     printf("%.1lf%\n", Nc / Nt*100.0);     }  system("pause");  return 0;}


评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月13日 10:25部分正确211063C++ (g++ 4.7.2)19436datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确138412/121答案正确13083/32答案正确11803/33答案正确194363/34运行超时  0/4
#include <iostream> #include <vector>#include<set>#include<iomanip>using namespace std;void readlnsets(vector<set<int>>*sets, int N){  int M, integers,index;  for (index = 0; index < N; index++)  {    cin >> M;    while (M--)    {      cin >> integers;      (*sets)[index].insert(integers);/*当数已经在集合里面就不再放入了*/    }   }}int main(){  int N,K,a,b;  double Nt, Nc;  set<int>Uset;  cin >> N;  vector<set<int>>sets(N);  readlnsets(&sets,N);  cin >> K;  while (K--)  {    cin >> a >> b;    a--;     b--;    Uset.insert(sets[a].begin(),sets[a].end());    Uset.insert(sets[b].begin(), sets[b].end());    Nt = Uset.size();    Nc = sets[a].size() + sets[b].size() - Nt;      cout << setiosflags(ios::fixed) << setprecision(1) << Nc / Nt *100.0<<"%"<< endl;    Uset.clear();  }  system("pause");  return 0;}