Wooden Sticks

题目描述

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).

 

标准输入

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.

 

标准输出

The output should contain the minimum setup time in minutes, one per line.

 

样例输入

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

 

样例输出

2

1

3

题意描述：以第一组样例为例，制作（4，9），（5,2），（2,1），（3,5），（1,4）5个木板。 例如（4,9）和（2,1），4大于2,9大于1，所以制作（2,1）浪费一分钟，当令一组数中的两个数都分别大于（2,1）时，制作这个木板将不需要时间。

所以（2,1），（5,2），（4，9）。（1,4），（3,5）.，所以需要两分钟。

（a,b）先按照a（或者b）进行排序，排序后根据b（或者a）分为多个集合。已经归入集合中的用flag [ ]进行标记。

#include
int main()
{
	int t,n,i,j,sum,h;
	int a[10001],b[10001],flag[10001];
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			flag[i]=0;
		sum=0;
		for(i=1;i<=n;i++)
			scanf("%d %d",&a[i],&b[i]);
		for(i=1;i=b[j])
				{
					flag[j]=1;
					b[i]=b[j];
				}
						
			}	
			sum=sum+1;
		}
		printf("%d\n",sum);
	}
	return 0;
}