POJ 1065 Wooden Sticks(LIS)

Wooden Sticks

Time Limit: 1000MS


Memory Limit: 10000K

Total Submissions: 20291


Accepted: 8556

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 ( 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


这个板子题真是各个OJ都有啊,题解请点击:http://blog.csdn.net/zwj1452267376/article/details/49981029


代码如下:


#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f
using namespace std;
int dp[5050];
struct node
{
	int l,w;
}a[5050];

int cmp(node a,node b)
{
	if(a.w==b.w)
		return a.l<b.l;
	return a.w<b.w;
}

int main()
{
	int n,t,i;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=0;i<n;++i)
			scanf("%d%d",&a[i].l,&a[i].w);
		sort(a,a+n,cmp);
		fill(dp,dp+n,INF);
		for(i=n-1;i>=0;--i)
			*lower_bound(dp,dp+n,a[i].l)=a[i].l;
		printf("%d\n",lower_bound(dp,dp+n,INF)-dp);
	}
	return 0;
}

 





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