Wooden Sticks

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

这也是一道贪心题，相当于让你砍树，怎么砍花费时间最少，砍一棵树需要花费分钟，但是如果下一颗树的长度和重量比当前这棵树要大的话，就不要时间。于是贪心的排序出现了，长度相同时重量小的放前面，长度不相同时，长度小的放前面。然后你做好出现不符合那个贪心规则的节点标记，就好了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int l;
    int w;
} a[50010];
int cmp(node p,node q)
{
    if(p.l==q.l)
        return p.w<q.w;
    else
        return p.l<q.l;
}
int main()
{
    int t,n,m,sum;
    int flag;
    scanf("%d",&t);
    while(t--)
    {
        int s[50010],j;
        scanf("%d",&n);
        for(int i=0; i<n; ++i)
            scanf("%d%d",&a[i].l,&a[i].w);
        sort(a,a+n,cmp);
        memset(s,0,sizeof(s));
        s[0]=1;
        m=0;
        sum=0;
        while(m<n)
        {
            ++sum;
            j=m;
            flag=1;
            for(int i=m+1; i<n; ++i)
            {
                if(s[i])
                    continue;
                if(a[j].l<=a[i].l&&a[j].w<=a[i].w)
                {
                    s[i]=1;
                    j=i;
                }
                else
                {
                    if(flag)
                    {
                        m=i;
                        flag=0;
                    }
                }
            }
            if(flag)
                break;
        }
        printf("%d\n",sum);
    }
    return 0;
}