Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10885 Accepted Submission(s): 4487

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

这个题因为我搜的题的时候作者就说了这是贪心的题目，我也没多想，就写了。反正对我来说能学到东西。

因为这个处理的中间过程的时候，还出错了。

思路用该很简单，就是先排序，按照长度，长度相同按宽度排序。

调用的c++函数，很快。

最好自己写排序函数，因为我都是调用函数，所以现在让我写快排什么的应该很费劲甚至写不出来。

长度有序了，那就不用管了。接下来我们只处理宽度就行了。

一遍一遍的遍历。遍历一遍就是结果+1，在遍历过程中，把这次能够处理的全部标记上已经处理了，下次就不用管了。

最后看看遍历了几次都处理了就行了。

贴上代码。

#include 
  
   
#include 
   
     #include 
    
      using namespace std; struct Point { int l; int w; }; bool comp(Point a,Point b ) { if(a.l!=b.l) { return a.l
     
      >cases; while(cases--) { int n; cin>>n; Point p[5001]; for(int i=0;i
      
       >p[i].l; cin>>p[i].w; } sort(p,p+n,comp); int m=0,t=0; bool b[5001]; int ans=0; memset(b,false,sizeof(b)); while(m
       
        =t) { m++; b[j]=true; t=p[j].w; } } } cout<
        
          代码简单易懂，我用结构体存储的木棒，用bool数组来标记是否处理过了。用m来表示当前已经处理多少元素了，如果处理元素和输入个数相同了，那么while结束。
          
         t就是当前处理的宽度，下次只能处理比当前这次宽的。
         好了！
         感谢自己坚持。

1051 Wooden Sticks贪心

Wooden Sticks