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：

 #include <iostream>

 using namespace std;
 struct stick{
     int l,w;
 }s[];
 int main()
 {
     int T;
     cin>>T;
     while(T--){
         int n;
         cin>>n;
         for(int i=;i<=n;i++)
             cin>>s[i].l>>s[i].w;
         //对l排序
         for(int i=;i<=n-;i++)
             for(int j=;j<=n-i;j++){
                 if(s[j].l>s[j+].l){
                     int t;
                     t=s[j].l;s[j].l=s[j+].l;s[j+].l=t;
                     t=s[j].w;s[j].w=s[j+].w;s[j+].w=t;
                 }
             }
         int sum=;  //递增序列的个数
         //对w序列进行筛选
         while(){
             int i;
             for(i=;i<=n;i++){  //如果全部为-1，则退出循环
                 if(s[i].w!=-)
                     break;
             }
             if(i>n) break;
             sum++;
             int num=;
             for(i=;i<=n;i++){
                 if(num== && s[i].w!=-){
                     num=s[i].w;
                     s[i].w=-;
                 }
                 else if(s[i].w!=- && s[i].w>=num){
                     num=s[i].w;
                     s[i].w=-;
                 }
             }
         }
         cout<<sum<<endl;

     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013