HDU1051(贪心)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8665    Accepted Submission(s): 3541

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
   
   
   
   

 

//先排个序，然后从第一个未访问的开始依次访问所有len,wei都大于或等于它的结点，重复进行直到结束，重复次数即为所求！

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

struct node
{
 int len,wei;
 bool flag;
}da[5000+10];

bool cmp(node a,node b)
{
 if(a.len!=b.len)
  return a.len<b.len;
 else
  return a.wei<b.wei;
}

int main()
{
 int i,t,n,k,st,ans;
 cin>>t;
 while(t--)
 {
  scanf("%d",&n);
  for(i=0;i<n;i++)
  {
   scanf("%d%d",&da[i].len,&da[i].wei);
   da[i].flag=false;
  }
  sort(da,da+n,cmp);

  ans=0;
  k=0;
  while(k<=n-1)
  {
   for(i=0;i<n;i++)
   {
    if(!da[i].flag)
    {
     da[i].flag=true;
     k++;
     ans++;
     st=i;
     break;
    }
   }

   for(i=st+1;i<n;i++)
   {
    if(!da[i].flag&&da[i].len>=da[st].len&&da[i].wei>=da[st].wei)
    {
     da[i].flag=true;
     k++;
     st=i;
    }
   }
  }
  printf("%d\n",ans);
 }
 return 0;
}