zoukankan      html  css  js  c++  java
  • Codeforces Round #407 (Div. 2) E The Great Mixing(BFS)

    Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration . Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration . The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.

    Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.

    Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration . Assume that the friends have unlimited amount of each Coke type.

    Input

    The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.

    The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.

    Output

    Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration , or -1 if it is impossible.

    Examples
    Input
    400 4
    100 300 450 500
    Output
    2
    Input
    50 2
    100 25
    Output
    3
    Note

    In the first sample case, we can achieve concentration using one liter of Coke of types and : .

    In the second case, we can achieve concentration using two liters of type and one liter of type: .

    题意:给你一些无限量不同浓度的二氧化碳,然后让你配出固定浓度为k的二氧化碳,问最少需要多少不同浓度的二氧化碳?

    思路:假设需要m个不同浓度的二氧化碳则m满足

    初始化x=0,求出下一次x=0时就得到了最后结果。

     1 #include <iostream>
     2 #include<cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 #include <queue>
     6 #include <map>
     7 using namespace std;
     8 queue<int>que;
     9 int a[1000005];
    10 map<int,int>mp;
    11 int main()
    12 {
    13     int n,m;
    14     scanf("%d%d",&n,&m);
    15     for(int i=0; i<m; i++)
    16     {
    17         scanf("%d",&a[i]);
    18         a[i]=a[i]-n;
    19     }
    20     sort(a,a+m);
    21     int cnt=unique(a,a+m)-a;
    22     que.push(0);
    23     while(!que.empty())
    24     {
    25         int x=que.front();
    26         que.pop();
    27         for(int i=0; i<cnt; i++)
    28         {
    29             int y=a[i]+x;
    30             if(abs(y)>=1001) continue;
    31             if(y==0)
    32             {
    33                 printf("%d
    ",mp[x]+1);
    34                 return 0;
    35             }
    36             if(!mp[y])
    37             {
    38                 mp[y]=mp[x]+1;
    39                 que.push(y);
    40             }
    41         }
    42     }
    43     printf("-1
    ");
    44     return 0;
    45 }
    View Code

    知识点:1.STL中unique是去重函数,2.在BFS的过程中使用map数组进行标记。

  • 相关阅读:
    ios 常用的数学函数
    iOS已发布应用中对异常信息捕获和处理
    iOS 制作framework步骤
    iOS 静态库和动态库.a和framework
    C#实现发送和接收pop3邮件方法
    C#实现给手机发送短信
    C#DataTable DataSet DataRow区别详解
    gridview合并相同的行
    ASP.NET导出EXCEl方法使用EXCEl对象
    ASP.NET导出EXCEl方法使用COM.EXCEL不使用EXCEl对象
  • 原文地址:https://www.cnblogs.com/wang-ya-wei/p/6686398.html
Copyright © 2011-2022 走看看