H - 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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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

题目大意：将木棍放在机器里处理，第一根需要一分钟，剩余的如果大于等于前边放入的长度和重量，就不用费时间，否则需要一分钟，计算给出一组数的最少时间！用贪心算出最少降序序列个数！

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
    int w,l;
}s[10000];
bool cmp(node x,node y)
{
    if(x.l==y.l)
        return x.w<y.w;
    return x.l<y.l;
}
int main()
{
    int t,i,n,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&s[i].l,&s[i].w);
        }
        sort(s,s+n,cmp);
        int s1,sum=0;
        for(i=0;i<n;i++)
        {
            s1=s[i].w;
            for(j=i+1;j<n;j++)
            {
                if(s1==0)
                    break;
                else if(s1<=s[j].w)
                {
                    s1=s[j].w;
                    s[j].w=0;
                }
            }
        }
        for(i=0;i<n;i++)
            if(s[i].w!=0)
            sum++;
        printf("%d\n",sum);
    }
    return 0;
}