B

题目描述：

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

213
本题的方法也是属于贪心算法，排序后把木头分组最后的组数就是最小的时间
代码：
#include<iostream>#include<vector>#include<algorithm>#include<iterator>#include<memory.h>using namespace std;struct wood{    int longth;    int weight;
};bool comp(const wood &m1,const wood  &m2){   if (m1.longth==m2.longth)    return m1.weight<m2.weight;   else if(m1.longth<m2.longth)    return true;   return false;}int main(){ int n;  cin>>n;  while(n>0)  {  n--;  int n2;  cin>>n2;  vector<wood>v;
  for(int i=0;i<n2;i++)  {      wood w;      cin>>w.longth>>w.weight;      v.push_back(w);  }   sort(v.begin(), v.end(), comp);   vector<wood>::iterator it=v.begin();   int a[5000];   memset(a,0,sizeof(a));   int k;   a[0]=1;   for(int i=1;i<n2;i++)   {   k=0;      for(int j=0;j<i;j++)        if(v[i].weight<v[j].weight&&k<a[j])        k=a[j];      a[i]=k+1;   }   int sum=0;   for(int i=0;i<n2;i++)   {if(sum<a[i])   sum=a[i];
   }   cout<<sum<<endl;  }
    return 0;
}
心得，一开始想到分组可是一直没想好分组后如何记录最大一开始采用单一变量的形式来存，后来发现实现起来困难，后来想了想可以反过来不一定要一次就记录最大的分组可以，把所有的木头的分组都记录下来然后输出最大值。