HDU1051:Wooden Sticks

Submit Status

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

题意是：将木棍放在机器里处理，第一根需要一分钟 下一根的长度和重量如果大于等于前边放入的长度和重量，就不用费时间，否则需要一分钟调试，计算给出一组数的最少时间！  

先用sort从小到大排序 算出非严格递增子序列的个数！  

换个思路也就是求最大递减子序列的长度 这里不过多解释 多思考 举几个例子就明白了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[10001];
struct node
{
	int x,y;
}s[10001];
bool cmp(node a,node b)
{
	if(a.x !=b.x )
	return a.x <b.x ;
	return a.y <b.y ;
}
int main()
{
	    int n,t;
	    scanf("%d",&t);
    while(t--)
    {
    	scanf("%d",&n);
    	for(int i=0;i<n;i++)
    	{
    		scanf("%d%d",&s[i].x ,&s[i].y );
    		dp[i]=1;
		}
		sort(s,s+n,cmp);
	
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<i;j++)
			{
				if(s[j].y >s[i].y &&dp[i]<dp[j]+1)
				{
					dp[i]=dp[j]+1;
				}
			}
		}
		int max=0;
		for(int i=0;i<n;i++)
		{
			if(dp[i]>max)
			{
				max=dp[i];
			}
		}
		printf("%d
",max);
	}
	return 0;
}