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). 
InputThe 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. 
OutputThe 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

題目大意：有一堆待加工的木棒，木棒有屬性長度L和質量W，我們對木棒加工有準備時間，規則如下：

1.第一根木棒花費準備時間1min

2.加工長度與質量均比上一根大的不需要準備時間，反之需要花費1min

要求求最小的準備時間。

按照題意分析，我們將L與W看做兩個數列，兩個數列遞增則不需要準備時間。即有幾個這樣雙遞增的數列就需要花費幾分鐘準備時間。

首先我們先找一個單遞增的數列（L或者W） 這裡我們以L為關鍵字進行排序。L相同的情況下按照W從小到大排序。

現在L已經是單調遞增數列。我們從頭遍歷，找有幾個單增的W數列（方法很簡單就是有一個變數記錄上一個木棒的W屬性，對於當前訪問的木棒i，若Wi<W，則是一個新的數列，否則continue）。新增標記陣列v，已加工的木棒標為true。 

原始碼：

#include<iostream>
#include<algorithm>
using namespace std;

struct mytype
{
	int length;
	int weight;
	bool vis;
};

int mycompare(mytype a,mytype b)
{
	if (a.length==b.length) return a.weight<b.weight;
	return a.length<b.length;
}

int main()
{
	int k;
	cin>>k;
	for (int o=1;o<=k;o++)
	{
		int n;
		mytype a[5010];
		cin>>n;
		for (int i=1;i<=n;i++) a[i].vis=false;
		for (int i=1;i<=n;i++) cin>>a[i].length>>a[i].weight;
		sort(a+1,a+n+1,mycompare);
		int ans=0;int wei;
		for (int i=1;i<=n;i++)
		{
			if (a[i].vis) continue;
			if (!a[i].vis)
			{
				ans++;
				a[i].vis=true;
				wei=a[i].weight;
			}
			for (int j=i+1;j<=n;j++)
			{
				if (a[j].vis) continue;
				if (!a[j].vis&&a[j].weight<wei) continue;
				if (!a[j].vis&&a[j].weight>=wei)
				{
					a[j].vis=true;
					wei=a[j].weight;
					continue;
				}
			}
		}
		cout<<ans<<endl;	
	}
	return 0;
}