zoukankan      html  css  js  c++  java
  • 1063. Set Similarity (25)

    题目如下:

    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%

    这道题目要求从两个集合中不重复的找到公共部分Nc和所有不重复元素Nt,然后计算Nc占Nt的比例。

    如果使用多个map进行查询,题目很容易完成,但是这道题对于滥用map进行了限制,我最初使用多个map,查找方式找出公共部分和全部不重复元素时最后一个case出现了超时,后来换了一个思路。

    只使用一个map,用于过滤在一个集合输入过程中重复的元素,这样得到的每个集合元素都是不同的,然后把每个集合元素按照升序排序,最后比较时就可以不同map了,我们使用两个指针curA、curB分别代表遍历到的集合A、集合B的位置。

    初始化Nc = Nt = 0,按照下面的操作进行:

    cur的移动原则:谁小移动谁去跟大的接近。

    ①检查curA和curB是否越界,越界则跳到③。

    如果setA[curA] > setB[curB],此时应该移动curB,根据移动原则,在curB移动过程中,所有遇到的setB[curB]元素均为setA中所没有的,此时Nt++;如果setA[curA] == setB[curB],说明碰到了公共元素,此时Nc++,Nt++(注意公共元素也算作不同元素中的一个),并且curA和curB均移动;如果setA < setB[curB],此时应该移动curA,根据移动原则,所有遇到的setA[curA]元素均为setB所没有的,此时Nt++。

    ③检查setA和setB是否有没遍历完的,如果有,说明剩下的都应该属于Nt。

    代码如下:

    #include <iostream>
    #include <map>
    #include <stdio.h>
    #include <vector>
    #include <algorithm>
    
    using namespace std;
    
    int main()
    {
        int Nc,Nt;
        map<int,int> commonMap;
        int N,M,ele;
        cin >> N;
        vector<vector<int> > sets(N+1);
        for(int i = 1; i <= N; i++){
            scanf("%d",&M);
            commonMap.clear();
            for(int j = 0; j < M; j++){
                scanf("%d",&ele);
                if(commonMap.find(ele) == commonMap.end()){
                    commonMap[ele] = 1;
                    sets[i].push_back(ele);
                }
            }
            sort(sets[i].begin(),sets[i].end());
        }
        int K,a,b;
        cin >> K;
        for(int i = 0; i < K; i++){
            scanf("%d%d",&a,&b);
            vector<int> setA = sets[a];
            vector<int> setB = sets[b];
            Nc = Nt = 0;
            int curA = 0;
            int curB = 0;
            while(curA < setA.size() && curB < setB.size()){
                if(setA[curA] < setB[curB]){
                    Nt++;
                    curA++;
                }else if(setA[curA] > setB[curB]){
                    Nt++;
                    curB++;
                }else{
                    Nc++;
                    Nt++;
                    curA++;
                    curB++;
                }
            }
            if(curA < setA.size()) Nt += setA.size() - curA;
            if(curB < setB.size()) Nt += setB.size() - curB;
            printf("%0.1f%%
    ",(float)Nc / Nt * 100);
        }
    
        return 0;
    }
    


  • 相关阅读:
    再谈 最速下降法/梯度法/Steepest Descent
    再谈 共轭方向法/Conjugate Direction Method In Optimization
    最速下降法/steepest descent,牛顿法/newton,共轭方向法/conjugate direction,共轭梯度法/conjugate gradient 及其他
    conjugate gradient method (共轭梯度法)
    牛顿法
    海森矩阵 Hessian matrix
    Jackknife,Bootstraping, bagging, boosting, AdaBoosting, Random forest 和 gradient boosting的区别
    机器学习算法中如何选取超参数:学习速率、正则项系数、minibatch size
    机器学习中的数学(3)-模型组合(Model Combining)之Boosting与Gradient Boosting
    无约束最优化方法
  • 原文地址:https://www.cnblogs.com/aiwz/p/6154097.html
Copyright © 2011-2022 走看看