1063 Set Similarity (25 分)

1063 Set Similarity (25 分)

 

Given two sets of integers, the similarity of the sets is defined to be /, 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 (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by M integers in the range [0]. After the input of sets, a positive integer K (≤) 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%

 

一个贪心过程，要求每次都是不一样的。

 

#include <bits/stdc++.h>

using namespace std;
int n,m,x,y,k;
vector<int> v[60];

int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &k);
        for(int j = 0 ; j < k; j++){
            scanf("%d",&m);
            v[i].push_back(m);
        }
        sort(v[i].begin(), v[i].end());
    }
    scanf("%d",&m);
    while(m--){
        scanf("%d%d", &x,&y);
        int same = 0, all = 0;
        int i = 0, j = 0;
        while(i < v[x].size() && j < v[y].size()){
            if(v[x][i] < v[y][j]){
                all++;
                i++;
            }else if(v[x][i] > v[y][j]){
                all++;
                j++;
            }else{
                all++;
                same++;
                i++;j++;
            }
            while(i &&i < v[x].size()&& v[x][i] == v[x][i-1]){
                i++;
            }
            while(j &&j < v[y].size()&& v[y][j] == v[y][j-1]){
                j++;
            }
        }
        while(i < v[x].size()){
            all++;
            i++;
            while(i &&i < v[x].size()&& v[x][i] == v[x][i-1]){
                i++;
            }
        }
        while(j < v[y].size()){
            all++;
            j++;
            while(j &&j < v[y].size()&& v[y][j] == v[y][j-1]){
                j++;
            }
        }
        // cout << same << " " << all <<endl;
        double ans = (same*1.0)/(all*1.0)*100.0;
        printf("%0.1lf", ans);
        cout<<"%"<<endl;
    }

    return 0;
}