hdu 1051:Wooden Sticks（水题，贪心）

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10423    Accepted Submission(s): 4287

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)

 

 

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　　水题。

　　先对木材长度 l 排序，在对 w 依次处理，将第一个递增序列筛选出来，筛选出来的元素赋值为-1，再将第二个递增序列筛选出来，同样标记-1 ，直到w序列全部为-1，代表筛选完成。输出记录下的递增序列的个数就是结果。

　　还可以用贪心做，有时间再做一遍。

　　代码：

 1 #include <iostream>

 2 

 3 using namespace std;  4 struct stick{  5     int l,w;  6 }s[5010];  7 int main()  8 {  9     int T; 10     cin>>T; 11     while(T--){ 12         int n; 13         cin>>n; 14         for(int i=1;i<=n;i++) 15             cin>>s[i].l>>s[i].w; 16         //对l排序

17         for(int i=1;i<=n-1;i++) 18             for(int j=1;j<=n-i;j++){ 19                 if(s[j].l>s[j+1].l){ 20                     int t; 21                     t=s[j].l;s[j].l=s[j+1].l;s[j+1].l=t; 22                     t=s[j].w;s[j].w=s[j+1].w;s[j+1].w=t; 23  } 24  } 25         int sum=0;  //递增序列的个数 26         //对w序列进行筛选

27         while(1){ 28             int i; 29             for(i=1;i<=n;i++){  //如果全部为-1，则退出循环

30                 if(s[i].w!=-1) 31                     break; 32  } 33             if(i>n) break; 34             sum++; 35             int num=0; 36             for(i=1;i<=n;i++){ 37                 if(num==0 && s[i].w!=-1){ 38                     num=s[i].w; 39                     s[i].w=-1; 40  } 41                 else if(s[i].w!=-1 && s[i].w>=num){ 42                     num=s[i].w; 43                     s[i].w=-1; 44  } 45  } 46  } 47         cout<<sum<<endl; 48 

49  } 50     return 0; 51 }

 

Freecode : www.cnblogs.com/yym2013