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2013
11-10

POJ 1862 Stripies [解题报告] Java

Stripies

问题描述 :

Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian – polosatiki, but the scientists had to invent an English name to apply for an international patent). The stripies are transparent amorphous amebiform creatures that live in flat colonies in a jelly-like nutrient medium. Most of the time the stripies are moving. When two of them collide a new stripie appears instead of them. Long observations made by our scientists enabled them to establish that the weight of the new stripie isn’t equal to the sum of weights of two disappeared stripies that collided; nevertheless, they soon learned that when two stripies of weights m1 and m2 collide the weight of resulting stripie equals to 2*sqrt(m1*m2). Our chemical biologists are very anxious to know to what limits can decrease the total weight of a given colony of stripies.

You are to write a program that will help them to answer this question. You may assume that 3 or more stipies never collide together.

输入:

The first line of the input contains one integer N (1 <= N <= 100) - the number of stripies in a colony. Each of next N lines contains one integer ranging from 1 to 10000 - the weight of the corresponding stripie.

输出:

The output must contain one line with the minimal possible total weight of colony with the accuracy of three decimal digits after the point.

样例输入:

3
72
30
50

样例输出:

120.000

解题代码:

import java.util.*;   
import java.text.*;   
  
public class Main {   
  
    public static void main(String[] args) {   
        Scanner cin = new Scanner(System.in);   
           
        int num = Integer.valueOf(cin.nextLine()).intValue();   
        List list = new ArrayList();   
           
        for(int i = 0; i < num; i++)   
            list.add(Double.valueOf(cin.nextLine()));   
           
        Collections.sort(list);   
           
        double result, temp = 0;   
        double a, b = 0;   
           
        if(list.size() == 1)   
            result = Double.valueOf((Double)list.get(0)).doubleValue();   
        else  
        {   
            int index = list.size() - 1;   
            a = Double.valueOf((Double)list.get(index)).doubleValue();   
            b = Double.valueOf((Double)list.get(index-1)).doubleValue();   
               
            result = 2 * Math.sqrt(a * b);   
               
            for(int i = index - 2; i>=0; i--)   
            {   
                a = Double.valueOf((Double)list.get(i)).doubleValue();   
                result = 2 * Math.sqrt(a * result);   
            }   
        }   
           
        DecimalFormat df = new DecimalFormat("#.000");   
        System.out.println(df.format(result));   
    }   
  
}

  1. 在方法1里面:

    //遍历所有的边,计算入度
    for(int i=0; i<V; i++)
    {
    degree = 0;
    for (j = adj .begin(); j != adj .end(); ++j)
    {
    degree[*j]++;
    }
    }

    为什么每遍历一条链表,要首先将每个链表头的顶点的入度置为0呢?
    比如顶点5,若在顶点1、2、3、4的链表中出现过顶点5,那么要增加顶点5的入度,但是在遍历顶点5的链表时,又将顶点5的入度置为0了,那之前的从顶点1234到顶点5的边不是都没了吗?

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