PAT-1063 Set Similarity （set集合）

Given two sets of integers, the similarity of the sets is defined to be Nc/Nt*100%, where Nc is the number of distinct common numbers shared by the two sets, and Nt 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%

题目大意：输入n个集合，每个集合中有若干数，现在需要做k次查询，每次给出要比较的两个集合，要求计算出相似度 = Nc / Nt * 100%，其中Nc是两个集合的交集的大小，Nt是两个集合并集的大小。

主要思路：考虑到每一个集合中可能存在重复的数，而且需要做大量的查找操作（找交集和并集时对集合a的每个元素都要判断是否存在于集合b），很容易想到stl库中的set容器，因为set中不存在重复元素，而且查找操作很快。对于每次查找操作，设置初始值nc为 0， nt初始值为集合b的大小，集合a的每个元素，如果存在于集合b，则nc+1；如果不存在，则nt+1。

（注意：如果用两集合大小之和 减去 两集合交集大小 的方法来计算 nt，可能会出现超时）

#pragma warning(disable: 4786)#include <cstdio>#include <vector>#include <set>using namespace std;int main(void) {    int n, i, j;        scanf("%d", &n);    vector<set<int> > vec(n);    set<int>::iterator iter;     int m, num;    for (i = 0; i < n; i++) {        scanf("%d", &m);        for (j = 0; j < m; j++) {            scanf("%d", &num);            vec[i].insert(num);        }            }    int k, a, b;    scanf("%d", &k);    for (i = 0; i < k; i++) {         scanf("%d%d", &a, &b);        int nc = 0, nt = vec[b-1].size();        for (iter = vec[a-1].begin(); iter != vec[a-1].end(); iter++) {            if (vec[b-1].count(*iter))                  //if (vec[b-1].find(*iter) != vec[b-1].end())                nc++;elsent++;        }//  nt = vec[a-1].size() + vec[b-1].size() - nc;//这样计算可能会超时        printf("%.1f%%\n", nc * 1.0 / nt * 100);           }        return 0;}

爬虫中的set容器解决这个问题就更容易了，& 和 | 分别对应交集和并集，唯一不足的就是有一个用例超时了。

n = int(input())L1 = []for i in range(n):    st = input()    L2 = st.split(' ')    L1.append(set(L2[1:]))k = int(input())for i in range(k):    pair = input().split(' ')    x, y = int(pair[0]), int(pair[1])    similarity = len(L1[x-1] & L1[y-1]) / len(L1[x-1] | L1[y-1]) * 100    print('%.1f%%' % (similarity))