首页 > 代码库 > hdu 1051 Wooden Sticks

hdu 1051 Wooden Sticks

2024-07-12 04:42:59   219人阅读

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11244    Accepted Submission(s): 4627


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
 

题目大意：
给n根木棍的长度和重量。根据要求求出制作木棍的最短时间。建立第一个木棍需要1分钟，若是接着要制作的木棍重量和长度都比此木棍长就不需要建立的时间，若是没有，则再需要建立时间。求时间最小为多少。

解题思路：
对木棍的长度和重量进行排序，以长度为首要考虑。排序完后的不一定都是下一根木棍重量和长度都大于前一根的。于是，我们对排序后的数组进行多次扫描，将可以在一次建立时间内完成的进行标记，知道木棍全部标记（设置一个外部变量来计数已扫描的元素的数量）。

例子：
5
4 9  5 2 2 1  3 5  1 4
排序完后：
1 4  2 1 3 5 4 9 5 2
进行第一次扫描：使用mark[]数组进行标记,mark[]初始化为0，红色为第一次描过的。
Stiks： (1 4)  (2 1)  (3 5)  (4 9)  (5 2)
Mark：    1       0       1       1      0
这是的setuptime为建立第一根木棍所要的时间，即1，此时扫描计数为3
接着进行第二次扫描，蓝色为第二次扫描过的结果。
Stiks： (1 4)  (2 1)  (3 5)  (4 9)  (5 2)
Mark：    1       0      1       1      0
这是的setuptime为1，此时扫描计数为5

代码如下：
#include<stdio.h>
#include<iostream>
#include<algorithm>
#define M 10000
using namespace std;

struct node
{
	int x,y,z;
}p[M];

int cmp(node a,node b)
{
	if(a.x<b.x) return 1;
	else 
	{
		if(a.x==b.x)
		{
			if(a.y<b.y) 
				return 1;
		}
		else return 0;
	}
	
}


int main ()
{ 
	int n,m;
	int i,j;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d",&m);
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&p[i].x,&p[i].y);
			p[i].z=0;
		}
		sort(p,p+m,cmp);
		int sum=0;
		for(i=0;i<m;i++)
		{
			if(p[i].z==0)
			{
				p[i].z=1;
				sum++;
				int k=p[i].y;
				for(j=i+1;j<m;j++)
				{
					if(p[j].z==0 && p[j].y>=k)
					{
						p[j].z=1;
						k=p[j].y;
					}
				}
			}
		}	
		printf("%d\n",sum);
	}
	return 0;
}


联系
我们