Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23191 Accepted Submission(s): 9411
Problem Description
There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:
(a) The setup time for the first wooden stick is 1 minute.
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup.
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.
Output
The output should contain the minimum setup time in minutes, one per line.
Sample Input
3 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1
Sample Output
2 1 3
Solution
//贪心问题,先处理最长最重的木头以便一台机器可以处理尽可能多的木头,
Code
/**
* date:2017.11.15
* author:孟小德
* function:杭电acm1051
*
*/
import java.util.*;
public class acm1051
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int num = input.nextInt();
int[] result = new int[num];
for (int j=0;j<num;j++)
{
int wood_num = input.nextInt();
wood[] array = new wood[wood_num];
for (int i=0;i<wood_num;i++)
{
int len = input.nextInt();
int weight = input.nextInt();
array[i] = new wood(len,weight);
}
sort(array);
int use_num = 0;
result[j] = 0;
wood temp = new wood();
while (use_num < wood_num)
{
for (int i=0;i<wood_num;i++)
{ //寻找未处理木头中最长最重的一根
if(array[i].use == false)
{
temp = array[i];
break;
}
}
for (int i=0;i<wood_num;i++)
{// 寻找可以被同时处理的木头
if (array[i].use == false && array[i].len <= temp.len &&
array[i].weight <= temp.weight)
{
array[i].use = true;
use_num++;
temp = array[i];
}
}
result[j]++;
}
}
for (int i=0;i<result.length;i++)
{
System.out.println(result[i]);
}
}
//按长度排序,长度相同则按重量排序
public static void sort(wood[] array)
{
boolean flag = true;
wood temp = new wood();
for (int i = 0;i<array.length && flag;i++)
{
flag = false;
for (int j = 0;j<array.length-i-1;j++)
{
if (array[j].len < array[j+1].len)
{
temp.len = array[j].len;
temp.weight = array[j].weight;
array[j].len = array[j+1].len;
array[j].weight = array[j+1].weight;
array[j+1].len = temp.len;
array[j+1].weight = temp.weight;
flag = true;
}
else if (array[j].len == array[j+1].len)
{
if (array[j].weight < array[j+1].weight)
{
temp.len = array[j].len;
temp.weight = array[j].weight;
array[j].len = array[j+1].len;
array[j].weight = array[j+1].weight;
array[j+1].len = temp.len;
array[j+1].weight = temp.weight;
flag = true;
}
}
}
}
}
}
class wood
{
public int len;
public int weight;
public boolean use = false;
public wood()
{
}
public wood(int len,int weight)
{
this.len = len;
this.weight = weight;
}
}
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