杭电ACM——处理木棍（贪心）

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

突破口：长度l与重量w等价，先将l按小到大排好序（选w也行一样的），接下从第一个开始遍历木棍，找出合适的木棍，下一个循坏起点，应为第一个不满足要求的木棍。

代码如下：

#include
#include
using namespace std;
struct stick
{
	int l;
	int w;
 }s[5005];
int fa[5005];  //可以一起处理的木棍视为一个集合，fa[]数组用来记录某根木棍所属的集合
bool cmp(stick x,stick y)
{
	if(x.l!=y.l)
	    return x.l=wei)
						{
							wei=s[j].w;
							fa[j]=i;
						}
					//	else  //这是之前想多了的操作
					//	{
					//		if(first)
					//		{
					//			flag=j;
					//			first=0;
					//		}
					//	}
					}
				}
			}
		}//for(i=1;i<=n;i++) printf("%d %d %d\n",s[i].l,s[i].w,fa[i]);
		printf("%d\n",sum);
	}
	return 0;
}

分析：
1.目标是尽可能减少setup的时间，则每次setup后应处理尽可能多的木棍→将木棍分为最少的集合；
2.首选l和w比较小的，接下来找到第一个l’>=l,w’>=w的，再接着找第一个l’’>=l’,w’’>=w’的，直到找不着了→将l或w从小到大排序，考虑对象由两个变为1个，简化问题，再接着按这种步骤找木棍；
3.处理过的木棍不能在处理，要做好标记。