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 1

2
1
3

Asia 2001, Taejon (South Korea)

/*贪心算法*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct wooden
{
   int length;
   int weight;
   int flag;
};

int cmp(const void *a,const void *b)
{
    struct wooden *c=(struct wooden *)a;//类型转换
    struct wooden *d=(struct wooden *)b;
    if(c->length==d->length)
    {
        return c->weight - d->weight;//按照weight大小排序 由小到大
    }
    else return c->length - d->length;//由小到大
}

int main()
{
    int num=0;
    int n=0;
    int i=0,j=0,k=0;
    int max=0;
    int time=0;
    struct wooden wood[5001];

    scanf("%d",&num);
    for(i=0;i<num;i++)
    {
       memset(wood,0,sizeof(struct wooden)*5001);//标志位清0
       scanf("%d",&n);
       for(j=0;j<n;j++)
       {
           scanf("%d%d",&wood[j].length,&wood[j].weight);
       }

        qsort(wood,n,sizeof(struct wooden),cmp);//全部按照从小到大排序
        time=0;
        for(j=0;j<n;j++)
        {
           if(wood[j].flag) continue;//等于1表示处理过了
           max=wood[j].weight;
           for(k=j+1;k<n;k++)
           {
               if(wood[k].weight>=max && !wood[k].flag )
               {
                   max=wood[k].weight;
                   wood[k].flag=1;
               }
           }
           time++;
         }
       printf("%d\n",time);
    }
    return 0;
}