HDU1051，贪心

　　　　　　　　　　　　Wooden Sticks

 

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

 英语菜，理解能力菜，所以只能用百度翻译，弄了好一大会才搞懂（惭愧啊，搜了题解，才懂了题意，但我没有看答案)。废话少说。

这个就是个 加工木材，然后只要l1<=l2,w1<=w2，就没必要再花一分钟了。

于是我一顿搞，结果连样例都没过（👎）无奈题解走起，才感觉有眼前一亮的感觉。每一组有两个值，l和w，由于我们只要后面的大于前面的，就可以继续工作，所以我们只要先把l排从小到大排好，只看w就可以了。

但这个根据贪心的思想，需要跳着来，所以我们需要对每一组进行标记，用完了就标记上下次就不能用了。然后每次把当前i的w记为最大值，然后往下遍历，看看谁比他大，比他大，那么就标记上已加工好，最大值更新。

第二个for用完后，ans才能加一，因为我们把比当前i的w都用完了。

下面上代码喽

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int b[10005];
struct node
{
	int l,w;
}st[10005];
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;
	cin>>t;
	while(t--)
		{
			memset(b,0,sizeof(b));
			int n;
			cin>>n;
			for(int i=0;i<n;i++)
				cin>>st[i].l>>st[i].w;
			b[0]=0;
			sort(st,st+n,cmp);
			int ans=0;
			for(int i=0;i<n;i++)　　　　　　//以当前为起点
				{
					if(b[i])
						continue;
					int max=st[i].w;
					for(int j=i+1;j<n;j++)　　　　//找到比当前i的w大的，加工掉（标记掉）
						{
							if(!b[j]&&max<=st[j].w)
								{
									max=st[j ].w;
									b[j]=1;
								}
						
						}	
					ans++;
				}
			cout<<ans<<endl;
		}
}

　　

记录自己的成长，加油吧！