HDOJ题目地址：传送门

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17453    Accepted Submission(s): 7121

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

题意：一个机器生产木头，最开始要一分钟，如果下一个木头比上一个木头短和轻，就需要一分钟的时间调试机器，

如果比上一个长和重就不需要时间调试，求最后的最短时间。

解题：

贪心+递减子序列（严格来说是非递增子序列）。按长度从大到小排序，长度相同，按重量从大到小排序。多次寻找递减子序列，即每次都把这一个序列的木头处理掉，即启动时间+1,并把被处理掉的木头做上标记。用一个变量记录被处理掉木头的个数，多次寻找子序列，处理掉木头，当该变量的个数等于木头总数时，退出循环。

比如第一个测试数据

4 9 5 2 2 1 3 5 14

排序后，为了方便看，我们竖着写。排序后：

l   w

5  2

4  9

3  5

2  1

1  4

有两个递减子序列：

5  2      2  1

4  9      3   5    1   4

#include<iostream>
#include<stdio.h>
#include<memory.h>
#include<algorithm>
using namespace std;
struct Node{
   int length,weight;
   int ok;//用来标记木头是否被处理过
}result[5010];
/**
   sort排序函数
*/
bool cmp(Node a,Node b){
    if(a.length>b.length){
        return true;
    }else if(a.length==b.length){
       if(a.weight>b.weight)
        return true;
       return false;
    }
    return false;
}
int main(){
   int n,m,i,index;
   cin>>n;
   while(n--){
      cin>>m;
      //录入木头的长度和重量
      for(i=0;i<m;i++){
           cin>>result[i].length>>result[i].weight;
           result[i].ok=0;
      }
       //排序
       sort(result,result+m,cmp);
       int sum=0;int index=0;//已经被处理的木头的个数
       Node temp;
        while(1){
            if(index==m)//当所有的木头都被处理时，结束循环
                break;
            //每次启动的时候都要寻找最长最终的木头
            for(int i=0;i<m;i++){
                if(!result[i].ok){
                        temp=result[i];
                         break;//别忘了这一句
                    }
            }
            for(int i=0;i<m;i++){
                //前提是该木头没有被处理，所以必须加上！wood[i].ok
                if(!result[i].ok&&result[i].length<=temp.length&&result[i].weight<=temp.weight){
                    result[i].ok=1;//该木头被处理
                    index++;
                    temp=result[i];//及时更换长度最长，重量最重的那根木头，及递减子序列的当前元素的下一个元素。
                }
            }
            sum++;//一次启动处理结束。
        }
        printf("%d\n",sum);
   }
}

参考博客：传送门