HDU - 1051 Wooden Sticks 贪心 动态规划

•                         Wooden Sticks

•      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
          Total Submission(s): 23527    Accepted Submission(s): 9551
  

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：将一堆木棍进行加工，如果后一个木棍比前一个的长度重量都小则不需要改动，否则将花费一分钟时间改动。最后输出总时间。

思路：可使用sort将木棍按长度从大到小排序，长度相等时按重量从大到小排。然后从第一个开始按照重量进行判断，循环求出结果。（在循环时，长度一定减小，所以只需关注重量）

#include<stdio.h>
#include<algorithm>
using namespace std;
struct woodn
{
	int l;
	int w;
	bool flag;
}wood[5005];

bool comp(struct woodn a,struct woodn b)
{
	if (a.l != b.l)
	{
		return a.l > b.l;
	}
	else
	{
		return a.w > b.w;
	}
}
int main()
{
	int T, n;
	scanf("%d", &T);
	while (T--)
	{
		int min_w=0; 
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
		{
			scanf("%d %d", &wood[i].l, &wood[i].w);
			wood[i].flag = false;
		}
		sort(wood, wood + n, comp);
		int num = 0;

		for (int i = 0; i < n; i++)
		{
			if (wood[i].flag)
			{
				continue;
			}
			min_w = wood[i].w;
			for (int j = i + 1; j < n; j++)
			{
				if (min_w >= wood[j].w&&!wood[j].flag)
				{
					min_w = wood[j].w;
					wood[j].flag = true;
				}
			}
			num++;
		}
		printf("%d\n", num);
	}
	return 0;
}