hdu 1051:Wooden Sticks（贪心）

Wooden Sticks

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 3

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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


这道题是贪心问题，思路是将木材分堆，每堆的后一个木材的长度l和重量w都比前一个的要大。按这样的规则分出几堆木材，那么最后就要多花几分钟准备。
下面是题目中给出的几组例子的分析：

Min　　　　　                 1      2   ==>  min=2

4　　9　　      1　　4     1

5　　2　　　　2      1     0     1

2　　1    =>   3　　5     1

3　　5　　      4　　9     1

1　　4　　      5　　2     0     1

 

Min                               1       ==>   min=1

2　　2       　  1　　1　  1  

1　　1   =>    2　　2     1

2　　2            2　　2     1

 

Min                        　　 1    2    3    ==>   min=2

1　　3     　　 1　　3     1  

2　　2    =>   2　　2     0    1

3　　1            3　　1     0    0   1

即先对l排序，然后根据规则对w进行贪心选择，选择出几个队列，就是几分钟。
上代码：

 1 #include <iostream>

 2 using namespace std;

 3 

 4 struct WoodenSticks{

 5     int l;

 6     int w;

 7 }w[5001];

 8 

 9 int main()

10 {

11     int f[5001];

12     int T;

13     cin>>T;

14     while(T--){

15         int n,i,j;

16         cin>>n;

17         //输入

18         for(i=1;i<=n;i++)

19             cin>>w[i].l>>w[i].w;

20         //从小到大排序。l相同就对w排序

21         for(i=1;i<n;i++)

22             for(j=1;j<=n-1;j++){

23                 if(w[j].l>w[j+1].l){      //当前木材的长度大于下一个木材的长度的话，调换位置

24                     int t1,t2;

25                     t1=w[j].l;t2=w[j].w;

26                     w[j].l=w[j+1].l;w[j].w=w[j+1].w;

27                     w[j+1].l=t1;w[j+1].w=t2;

28                 }

29                 else if(w[j].l==w[j+1].l)   //当前木材的长度等于下一个木材的长度的话，比较木材的重量，如果当前大于下一个的，调换位置。

30                     if(w[j].w>w[j+1].w){

31                         int t1,t2;

32                         t1=w[j].l;t2=w[j].w;

33                         w[j].l=w[j+1].l;w[j].w=w[j+1].w;

34                         w[j+1].l=t1;w[j+1].w=t2;

35                     }

36             }

37 

38 

39         int min=1;

40         f[1]=w[1].w;

41         for(i=2;i<=n;i++){

42             for(j=1;j<=min;j++){

43                 if(w[i].w>=f[j]){

44                     f[j]=w[i].w;     //存储当前各队列最大的数

45                     break;

46                 }

47             }

48             if(j>min){  //如果小于前面每一个数，则重新开一个队列，min+1

49                 min++;

50                 f[min]=w[i].w;

51             }

52         }

53         cout<<min<<endl;

54     }

55     return 0;

56 }