贪心算法

POJ1065总结

Wooden Sticks

题目来源

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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

我的代码

按照重量排序后求长度的最大下降子序列的长度,所得就是结果。

#include<stdio.h>
#include<algorithm>

using namespace std;

struct stick
{
	int w,l;
};

bool cmp(stick a,stick b)
{
	if (a.w==b.w)	return a.l<b.l;
	else return a.w<b.w;
}


int main()
{
	int t;
	scanf("%d",&t);
	while (t--)
	{
		int n;
		int temp=1;
		int dp[5001];
		scanf("%d",&n);
		stick sticks[5000];
		for (int i=0;i<n;i++)
		{
			scanf("%d%d",&sticks[i].l,&sticks[i].w);
		}
		sort(sticks,sticks+n,cmp);
		for (int i=0;i<n;i++)
		{
			dp[i]=1;
			for (int j=0;j<i;j++)
			{
				if (sticks[i].l<sticks[j].l) dp[i]=dp[i]>dp[j]+1?dp[i]:dp[j]+1;
			}
			temp=dp[i]>temp?dp[i]:temp;
		}
		printf("%d\n",temp);
	}
	return 0;
}

反思

狄尔沃斯定理(Dilworth’s theorem)

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