1063. Set Similarity (25)

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%

解题思路：

输入数组后，根据输入的数组序号对数组进行排序，然后去重，再进行类似于merge操作的遍历，获得两个数组中不同的数的个数。

#include <iostream>
#include"stdio.h"
#include"stdlib.h"
#include"algorithm"
using namespace std;


struct store{
    int a[10000];
    int length;
    bool sort_flag;//标记数组是否已经排序，可以避免重复排序
    int cnt;
};
int main()
{
    int N,M,K;
    struct store s[50];
    scanf("%d",&N);
    for(int i=0;i<N;i++){
        scanf("%d",&M);
        s[i].length = M;
        s[i].cnt=M;
        s[i].sort_flag = false;
        for(int j=0;j<M;j++){
            scanf("%ld",&s[i].a[j]);
        }
    }
    scanf("%d",&K);

    int first,second;
    for(int k=0;k<K;k++){
        scanf("%d%d",&first,&second);
        first--;
        second--;
        if(!s[first].sort_flag){
            sort(s[first].a,s[first].a+s[first].length);//快排

            //去重
            for(int i=s[first].length-1;i>=1;i--){
                if(s[first].a[i]==s[first].a[i-1]){
                    s[first].a[i]=-1;
                    s[first].cnt--;//记下数组中distinct的数的个数
                }
            }
            s[first].sort_flag=true;//已经排序了
        }


        if(!s[second].sort_flag){
            sort(s[second].a,s[second].a+s[second].length);
            for(int i=s[second].length-1;i>=1;i--){
                if(s[second].a[i]==s[second].a[i-1]){
                    s[second].a[i]=-1;
                    s[second].cnt--;
                }
            }
            s[second].sort_flag=true;
        }
        int distinct=0;
        int i=0,j=0;
        while(i<s[first].length&&j<s[second].length){
            if(s[first].a[i]!=-1&&s[second].a[j]!=-1){
                if(s[first].a[i]<s[second].a[j]){
                    distinct++;
                    i++;
                }
                else if(s[first].a[i]>s[second].a[j]){
                    distinct++;
                    j++;
                }
                else{
                    i++;
                    j++;
                }
            }else{
                if(s[first].a[i]==-1&&s[second].a[j]!=-1){
                    i++;
                }else if(s[first].a[i]!=-1&&s[second].a[j]==-1){
                    j++;
                }
                else /*if(s[first].a[i]!=-1&&s[second].a[j]!=-1)*/{//用else if会超时
                    i++;
                    j++;
                }
            }
        }
        while(i<s[first].length){
            if(s[first].a[i]!=-1)
                distinct++;
            i++;
        }
        while(j<s[second].length){
           if(s[second].a[j]!=-1)
                distinct++;
            j++;
        }

        long common = (s[first].cnt+s[second].cnt-distinct)/2;
        printf("%.1f%\n",(common*100.0/(common+distinct)));
    }
    return 0;
}