PAT (Advanced Level) Practise 1063 Set Similarity (25)

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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%
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;}

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