Wooden Sticks

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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

Output for the Sample Input

2
1
3 

题目大意：

给你 N 个木棍，每根木棍都有一个长度 l 和一定重量 w 。

      机器每对一根进行操作需要一分钟的时间 ，但是如果当前进行操作的木棍的长度和重量都不小于前一根，

      那么这根木棍就不需要加时间。

算法：贪心 + 标记。

思路：先按照长度由小到大排序，如果长度一样，则按照重量由小到大排序。

      依次操作每根木棍，同时标记已经操作。

      操作当前木棍时，同时检查是否可以使它后面的木棍直接操作而且不用加时间，如果可以，则直接操作并且记

参考代码：

#include<stdio.h>#include<algorithm>using namespace std;struct ss{int l;int w;int ok;}note[5010];//先定义一个结构体来储存长度，重量，还有标记bool cmp(ss x,ss y){if(x.l==y.l)return x.w<y.w;return x.l<y.l;}//比较函数的定义；int main(){int t,n,i,j,k,f,m,a,b;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&note[i].l,&note[i].w); note[i].ok=0;}sort(note,note+n,cmp);k=0;//花费的时间for(i=0;i<n;i++){if(note[i].ok) continue;//保证之前的都标记过k++;note[i].ok=1;a=note[i].w;for(j=i+1;j<n;j++)//保证每次执行这个函数的时候，都扫描i以后的{if(note[j].ok==0&note[j].w>=a)//因为长度已经排好，所以这里之比较重量{a=note[j].w;//若满足条件更新a的值note[j].ok=1;}}}printf("%d\n",k);}return 0;}