Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18745    Accepted Submission(s): 7692

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

 

题意：

 有一个木棒处理机，每次处理一个木棒，处理的一个木棒时需要1分钟step时间，处理完一个木棒后下一次只能处理l和w都不小于此木棒的木棒，问最少的step时间。

代码：

1 /* 2 按照l由小到大排序，用一个数组存储机器不需要step的w值，如果下一个w大于等于此数组中的某一个用它更新与他差值 3 最小的那个，如果大于每一个数组值就再开一个数组来记录他。 4 */ 5 #include
      
        6 #include
       
         7 #include
        
          8 #include
         
           9 #include
          
           10 #include
           
            11 #include
            
             12 #include
             
              13 #include
              
               14 #include
               
                15 using namespace std;16 int t,n;17 int a[5010];18 struct stick19 {20 int l,w;21 }sti[5010];22 bool cmp(stick x,stick y)23 {24 if(x.l==y.l)25 return x.w
                
                 =a[j])&&(sti[i].w-a[j]