Wooden Sticks

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
  

 

Source

Asia 2001, Taejon (South Korea) 

分析 ：此题是典型的贪心算法。

主要 注意一点，比较时是相邻两个依次比较，而不是找到最小的一个与所有的比较。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;
struct wood{
	int len,weg;
}a[100000];
int cmp(wood b,wood c)
{
	if(b.len!=c.len)
	return b.len<c.len;
	else return b.weg<c.weg;
}
int main()
{
	int n,i,j,k,num,sum,t,count;
	int v[10010];
	scanf("%d",&n);
	while(n--)
	{
		memset(v,0,sizeof(v));
		scanf("%d",&k);
		for(i=0;i<k;i++)
		scanf("%d%d",&a[i].len,&a[i].weg);
		sort(a,a+k,cmp);
		int count=0;
		//for(i=1;i<k;i++)
		//if(a[i].weg<num)
		//{
		////	sum+=1;
			//num=a[i].weg;
		//}           //题目是相邻的两个进行比较 
		 for(i=0;i<k;i++)
		 {
		 	if(!v[i])
		 {
		 	v[i]=1;
		 	for(j=i+1,t=a[i].weg;j<k;j++)
		 	if(a[j].weg>=t&&(!v[j]))
		 	{
		 		t=a[j].weg;
		 		v[j]=1;
		 	}
		 	count++;
		 }
		 } 
		printf("%d\n",count);
	}
	return 0;
}