POJ1065-Wooden Sticks

最新推荐文章于 2021-02-07 23:59:09 发布

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最新推荐文章于 2021-02-07 23:59:09 发布

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分类专栏： POJ ----贪心

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本文介绍了一道关于木棍加工顺序的问题，旨在寻找最短加工准备时间的解决方案。通过将木棍按长度和重量排序，并采用贪心算法来确定最优处理顺序。

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Wooden Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23228		Accepted: 9979

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

Source

Taejon 2001

题意：要处理一些木棍，第一根的时间是1分钟，若后面一根木棍的长度和重量都大于等于前面一根的长度和重量，就不需要时间，否则需要1分钟，问如何安排处理木棍的顺序，才能使花的时间最少。

解题思路：贪心，把木棍先按照长度从小打到排序，若长度相同则按重量从小到大排序

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <cmath>  
#include <map>  
#include <cmath>  
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  

using namespace std;

#define LL long long  
const int INF = 0x3f3f3f3f;

int n,visit[5005];
struct node
{
	int l, w;
}x[5005];

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

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%d%d", &x[i].l, &x[i].w);
		sort(x + 1, x + 1 + n,cmp);
		int sum = 0;
		memset(visit, 0, sizeof visit);
		for (int i = 1; i <= n; i++)
		{
			if (visit[i]) continue;
			visit[i] = 1;
			int k = x[i].w;
			for (int j = i + 1; j <= n; j++)
				if (!visit[j] && k <= x[j].w)
					visit[j] = 1, k = x[j].w;
			sum++;
		}
		printf("%d\n", sum);
	}
	return 0;
}