1065 Wooden Sticks

Wooden Sticks
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 11769
 
Accepted: 4803
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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .
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 2n 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
Taejon 2001
 
//대수제#include#include#includeusingnamespacestd;struct T { int w,l; }a[5010]; bool cmp(T a,T b) { if(a.l==b.l) return a.w=l&a[i].w>=w) {w=a[i].w;l=a[i].l;vis[i]=1; tt++; } else if(!vis[i]&&!flag) { flag=true; num=i; } count++; } printf("%d/n",count); } return 0; }