PAT (Advanced Level) Practise 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:

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%
统计一下每个的组合各自的不同数字数目，然后排序，对于不同的之间找出相同的数字数量。
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,x,y;
int a[50][10000],c[50];

int main()
{
  scanf("%d",&n);
  for (int i=0;i<n;i++)
  {
    scanf("%d",&m);
    while (m--) scanf("%d",&a[i][c[i]++]);
    sort(a[i],a[i]+c[i]);
    c[i]=unique(a[i],a[i]+c[i])-a[i];
  }
  scanf("%d",&n);
  while (n--)
  {
    scanf("%d%d",&x,&y);
    x--;  y--;
    int com=0;
    for (int i=0,j=0;i<c[x]&&j<c[y];)
    {
      if (a[x][i]==a[y][j]) i++,j++,com++;
      else a[x][i]<a[y][j]?i++:j++;
    }
    printf("%.1lf%%\n",100.0*com/(c[x]+c[y]-com));
  }
  return 0;
}