练习1-b

编号： 1001 Problem B

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

213

题意 :  加工木头，木头有长度和重量之分，每一次加工木头需要1分钟时间进行准备，如果下一次加工的木头的长度和重量均大于上次加工的木头的长度和重量，则不需要时间进行准备。求加工所有木头所需的最短时间。

思路：第一次想着写一个排序函数，直接按照最优解的加工顺序把各组数据排列出来，结果证明是我错了，后来又咨询

了一下远飞，才知道这题的做法个上一题是非常相似的，还是用安排活动的算法进行解题，木头的长度和重量相当于活动的起始和终止时间，对长度进行升序排列后，然后对重量进行多次的筛选，然后就可得出结果。

感想：在做完第一题后，对贪心算法的理解加深了不少，然后这一题做起来就容易多了。然后第三题做起来就不是那么容易了，加油！

代码：

# include <iostream># include <vector># include <algorithm># include <fstream>using namespace std;struct wooden{    int l;    int w;    int nu;    bool operator < (const wooden &a) const    {            return l <= a.l ;    }};int main(){    //ifstream cin ("b.txt");    vector<wooden> wo;    wooden wwo;    int n;    cin >> n;    while(n--)    {        wo.clear();        int m;        cin >> m;        while (m--)        {            cin >> wwo.l >> wwo.w;            wo.push_back(wwo);        }        sort (wo.begin(), wo.end());        //for (int i = 0; i < wo.size(); i++)           // cout << wo[i].l <<" " << wo[i].w<<endl;        int time = 0;        for (int i = 0; i < wo.size(); i++)        {            if (wo[i].nu == 0)                continue;            time ++;            int count = i;            wo[i].nu = 0;            for (int j = 0; j < wo.size(); j++)            {                if (wo[count].w <= wo[j].w && wo[j].nu != 0)                {                    wo[j].nu = 0;                    count = j;                }            }        }        cout << time <<endl;    }    return 0;}