pat 1063

最新推荐文章于 2025-08-13 16:01:58 发布

原创最新推荐文章于 2025-08-13 16:01:58 发布 · 593 阅读

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此博客介绍了一种计算两个集合相似度的方法，通过比较共享元素的数量和总元素数量来得出百分比形式的结果。

1063. Set Similarity (25)

时间限制

300 ms

内存限制

32000 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%

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int num[51][10001];
int sm[51], tmp[20010];

void outPut(int a, int b)
{
	int nc = 0, nt = 0, nn, i;

	for(i = 0; i < sm[a]; i++)
		tmp[i] = num[a][i];

	for(; i < sm[a] + sm[b]; i++)
		tmp[i] = num[b][i-sm[a]];

	sort(tmp, tmp + sm[a] + sm[b]);

	for(i = 0; i < sm[a] + sm[b]; i++)
	{
		if(i == 0)
		{
			nt++;
			nn = 1;
		}
		else
		{
			if(tmp[i] != tmp[i-1])
			{
				nt++;
				nn = 1;
			}
			else if(tmp[i] == tmp[i-1])
			{
				nn++;
				if(nn == 2)
					nc++;
			}
		}
	}

	printf("%.1lf", (1.0 * nc) / (1.0 * nt) * 100);
	printf("%%\n");
}
int main()
{
	int i, j, a, b, cc, tt;
	int n, m;

	while(scanf("%d", &n) != EOF)
	{
		for(i = 0; i < n; i++)
		{
			scanf("%d", &sm[i]);

			for(j = 0, a = 0; j < sm[i]; j++)
			{
				scanf("%d", &tt);
				for(cc = 0; cc < a; cc++)
					if(num[i][cc] == tt) //将each of sets 中的same data delete
						break;

				if(cc == a)
					num[i][a++] = tt;
			}

			sm[i] = a;
		}

		scanf("%d", &m);

		for(i = 0; i < m; i++)
		{
			scanf("%d%d", &a, &b);
			outPut(a-1, b-1);
		}
	}
	return 0;
}