HDU1051--Wooden Sticks

水深无声 2022-08-23 12:54 19阅读 0赞

# Wooden Sticks #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 11070    Accepted Submission(s): 4553**

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

解析：简单的贪心问题，但是有点非常容易错误，每个stick只能减免一个的stick的执行时间，所以每次的数据需要更新，方法直接排序，可以按照长度或者重量排序都可以，对每个stick进行标记看是否执行过，然后从头开始遍历，对于每个stick看能减肥几个，没减免一个就需要更新其比较的长度或者重量，如果还是不清楚的话，可以看我的代码哈！

贴一下自己的代码哈！

#include<iostream>
    #include <algorithm>
    using std::endl;
    using std::cin;
    using std::cout;
    using std::sort;
    const int MAXN = 5000 + 200;
    struct stick{
    	int length;
    	int weight;
    	//标记是否访问过
    	int tag;
    }sticks[MAXN];
    //排序函数
    bool cmp(stick a , stick b)
    {
    	if(a.length > b.length)
    		return true;
    	if(a.length == b.length)
    		return a.weight > b.weight;
    	return false;
    }
    int main()
    {
    #ifdef LOCAL
    	freopen("input.txt" , "r" , stdin);
    #endif
    	int T;
    	cin >> T;
    	while(T--)
    	{
    		int n;
    		cin >> n;
    		for(int i=0; i<n; ++i)
    		{
    			cin >> sticks[i].length >> sticks[i].weight;
    			sticks[i].tag = 0;
    		}
    		sort(sticks , sticks+n , cmp);
    		int time = 0 , weight;
    		for(int i=0; i<n; ++i)
    		{
    			//对每个未访问过的stick进行遍历
    			if(!sticks[i].tag)
    			{
    				sticks[i].tag = 1;
    				time++;
    				weight = sticks[i].weight;
    				for(int j=i+1; j<n; ++j)
    				{
    					if(!sticks[j].tag && weight >= sticks[j].weight)
    					{
    						//如果符合条件可以减免然后更新下次比较的重量
    						weight = sticks[j].weight;
    						//更新访问标记
    						sticks[j].tag = 1;
    					}
    				}
    			}
    		}
    		cout << time << endl;
    	}
    	return 0;
    }