【HDU】1051 - Wooden Sticks（贪心 & 二分 || 动态规划）

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Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18101    Accepted Submission(s): 7387

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)

 

和上一道题差不多，不过这次是升序，g 数组里的数肯定是降序排列的，用lower_bound ( )就不会用了，手写了二分也能过。

这样是用贪心的方法做的。最下面贴出 LIS 的做法。

代码如下：

#include <cstdio>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int ans;
struct wood
{
	int l,w;
}data[5000+11];
bool cmp(wood a , wood b)
{
	if (a.l == b.l)
		return a.w < b.w;
	return a.l < b.l;
}
int g[5000+11];
int find(int x)
{
	int l = 1;
	int r = ans;
	int mid;
	while (r >= l)
	{
		mid = (l + r) >> 1;
		if (g[mid] > x)		//二分的判断条件还是有点晕 
			l = mid + 1;
		else
			r = mid - 1;
	}
	return l;
}
int main()
{
	int u;
	int n;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d",&n);
		for (int i = 1 ; i <= n ; i++)
		{
			scanf ("%d %d",&data[i].l,&data[i].w);
			g[i] = INF;
		}
		sort (data + 1 , data + 1 + n , cmp);
		ans = 0;
		int pos;
		for (int i = 1 ; i <= n ; i++)
		{
			pos = find(data[i].w);
			g[pos] = data[i].w;
			ans = max (ans , pos);
		}
		printf ("%d\n",ans);
	}
	return 0;
}