Wooden Sticks


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

题解：

这道题目是先要排序的，按照长度或者重量排都可以，当长度（重量）相同时就按照重量（长度）排，从大到小或从小到大都可以！这里我懂的，没有问题！
排序之后，问题就可以简化，（假设按照长度不等时长度排，长度等是按照重量排，我假设按照从大到小来排！）即求排序后的所有的重量值最少能表示成几个集合。长度就不用再管了，从数组第一个数开始遍历，只要重量值满足条件，那么这两个木棍就满足条件！
  
 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<vector>
 8 using namespace std;
 9 struct wooden{  
10     int l,w;  
11 };  
12 wooden my[5010];  
13 bool comp(wooden a,wooden b){  
14     if(a.l>b.l)return 1;  
15     else if(a.l==b.l)  
16         return a.w>b.w;  
17     else return 0;  
18 }  
19 int main()
20 {
21     int t;  
22     scanf("%d",&t);
23     while(t--){  
24         int n;  
25         scanf("%d",&n); 
26         int i=0,j;  
27         while(i<n)
28         {
29             scanf("%d %d",&my[++i].l,&my[i].w);
30         }
31         sort(my,my+n,comp);  
32         int out=n;  
33         for(i=1;i<n;i++)  
34             for(j=0;j<=i-1;j++){  
35             if(my[j].l>=my[i].l&&my[j].w>=my[i].w){  
36             out--;  
37             my[j].l=my[i].l;  
38             my[j].w=my[i].w;  
39             my[i].l=0;  
40             my[i].w=0;  
41             break;  
42             }  
43         }  
44        printf("%d\n",out);
45     }  
46     return 0;  
47 }