hdoj-1051

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18610    Accepted Submission(s): 7605

Problem Description

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
  
  

 

Source

Asia 2001, Taejon (South Korea)

感觉是贪心算法，先排序，然后不断地更新最大值，就行了，排序按照长度从小到大，长度相同时
按照重量从小到大，找一个数组存放每次出现的不能加入的值，可以加入就更新最大值

#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
	int l;
	int w;
}arr[6000];
int r[6000]; 
bool cmp(node a,node b)
{
	if(a.l==b.l)
		return a.w<b.w;
	return a.l<b.l;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int i,j;
		for(i=0;i<n;++i)
			scanf("%d%d",&arr[i].l,&arr[i].w);
		sort(arr,arr+n,cmp);
		int sum=1;
		r[0]=arr[0].w;
		for(i=1;i<n;++i)
		{
			for(j=0;j<sum;++j)
			{
				if(arr[i].w>=r[j])//更新最大值 
				{
					r[j]=arr[i].w;
					break;
				}
			}
			if(j==sum)//没有找到比它小的 
			{
				r[sum]=arr[i].w;
				++sum;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}