1063?Set Similarity?(25?分)

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?Nlines follow, each gives a set with a positive?M?(≤10?4??) and followed by?M?integers in the range [0,10?9??]. 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%

題意：集合相似性=（兩個集合的公共數對/兩個集合互相不同的總數）*100%

求這個相似性

解析：排序二分查找，發現會有測試點超時，去除重復，用set

#include<bits/stdc++.h> using namespace std;#define e exp(1) #define pi acos(-1) #define mod 1000000007 #define inf 0x3f3f3f3f #define ll long long #define ull unsigned long long #define mem(a,b) memset(a,b,sizeof(a)) int gcd(int a,int b) {return b?gcd(b,a%b):a; }set<int> lis[3000]; int n,k;int main() {int no = 1;scanf("%d",&n);for(int i = 1; i <= n ; i++) {int m,v;scanf("%d",&m);for(int j = 0 ; j<m; j++) {scanf("%d",&v);lis[i].insert(v);}}scanf("%d",&k);for(int i =0 ; i < k ; i++) {int l1,l2,same = 0;scanf("%d%d",&l1,&l2);set <int> ::iterator it;for(it = lis[l1].begin() ; it!=lis[l1].end(); it++) {if(lis[l2].find(*it) != lis[l2].end())same++;}float rate = float(same) / float(lis[l1].size()+lis[l2].size()-same) * 100;printf("%.1f%%\n",rate);}return 0; }

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