题目描述

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 Output

2
1
3

题目大意

将n根木头放入一台机器加工,若要放的木头的长度与宽度都分别大于或等于上一根的长度与宽度,则需要调整机器。第一根放之前也需要调整机器。求需要调整几次

大致思路

贪心问题。贪心标准为一次调整加工的木头尽可能的多。因此需要第一根木头尽可能的小,然后找出所有与它相容的木头。所求调整次数即为最优解。

下面的代码使用了数组(使用容器会方便一些,当时有些偷懒。。。。)先由小到大sort,从第一个开始,每当找到上一个相容的即从数组中除去(重新赋值为-1,使数据无效)寻找到最末,然后从下一个有效数据开始重复以上过程,知道数组中全为无效数据,执行此过程的次数即为解。

AC代码

  1. #include<iostream>
  2. #include<algorithm>
  3. #include<stdio.h>
  4. using namespace std;
  5. struct woods
  6. {
  7. int l;
  8. int w;
  9. }a[5000];
  10. bool cmp(woods a, woods b)
  11. {
  12. if (a.l < b.l)
  13. return 1;
  14. else if (a.l == b.l)
  15. {
  16. if (a.w < b.w)
  17. return 1;
  18. }
  19. return 0;
  20. }
  21. int main()
  22. {
  23. //freopen("date.in", "r", stdin);
  24. //freopen("date.out", "w", stdout);
  25. int N,m,sum,flag,count;
  26. cin >> N;
  27. woods tem;
  28. for (int i = 0; i < N; i++)
  29. {
  30. sum = 0;
  31. flag = 1;
  32. cin >> m;
  33. for (int j = 0; j < m; j++)
  34. {
  35. cin >> a[j].l >>a[j].w;
  36. }
  37. sort(a, a+m, cmp);
  38. /*for (int l = 0; l < m; l++)
  39. {
  40. cout << a[l].l << " " << a[l].w << endl;
  41. }*/
  42. for (int k = 0; k < m ; k++)
  43. {
  44. //count = 0;
  45. tem.l = a[k].l;
  46. tem.w = a[k].w;
  47. if (a[k].l != -1)
  48. {
  49. sum++;
  50. for (int l = k; l < m; l++)
  51. {
  52. if (a[l].l >= tem.l&&a[l].w >= tem.w)
  53. {
  54. tem.l = a[l].l;
  55. tem.w = a[l].w;
  56. a[l].l = -1;
  57. a[l].w = -1;
  58. //count++;
  59. }
  60. }
  61. }
  62. }
  63. cout << sum<<endl;
  64. }
  65. }

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