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).

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.

The output should contain the minimum setup time in minutes, one per line.

       

        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 
       

       

        2
1
3
       

Asia 2001, Taejon (South Korea)

We have carefully selected several similar problems for you:  1045 1053 1789 1257 1003 

题目大意+解析：

有许多木头，木头有两个属性，重量和高度，有若干机器，机器有启动时间，启动时间为1minute，有个规则是机器切割木头，如果第二根木头比第一根木头的宽高都大的话，机器没有启动时间，请问机器最小启动时间是多少。简单的贪心，排列木头的顺序，使的机器在一次切割时可以尽可能多的切割木头，减少启动时间；思路get没，接下来就是代码。

#include<iostream>
using namespace std;
struct Strick{double width;double length;
}temp;
int main()
{int n,m,count,k;cin>>n;int i,j;for(i=0;i<n;i++){cin>>m;int flag;count=0;Strick *ps=new Strick[m];for(j=0;j<m;j++)cin>>ps[j].length>>ps[j].width;for(k=0;k<m-1;k++)	for(j=0;j<m-1-k;j++)if(ps[j].length>ps[j+1].length){temp=ps[j];ps[j]=ps[j+1];ps[j+1]=temp;}	for(int x=0;x<m;x++){	if(ps[x].width){k=ps[x].width;for(j=0;j<m;j++)if(k<=ps[j].width){k=ps[j].width;ps[j].width=0;flag=1;}if(flag)count++;}}cout<<count<<endl;delete[] ps;ps=NULL;	}return 0;
}