HDU 1051(贪心)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1051

 

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13534    Accepted Submission(s): 5590

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

 

 

题目大意：一堆木棍，有长度和重量，处理第一根需要一分钟，然后处理下一根木棍，如果这根木棍的长度和重量都不大于之前的木棍，则不需要花费时间，否则也需要一分钟。

分析：首先排序，先按长度排序，如果长度一样，则按重量排序（从小到大和从大到小排序都可以，我是按从小到大排序的）。然后求递增（减）序列的最小个数。

 

 1 #include <cstdio>

 2 #include <cmath>

 3 #include <cstring>

 4 #include <iostream>

 5 #include <algorithm>

 6 using namespace std;

 7 

 8 int T,n;

 9 int a[5005],b[5005],vis[5001];

10 

11 int main ()

12 {

13     int i,j;

14     int time;

15     int x,y;

16     scanf ("%d",&T);

17     while (T--)

18     {

19         scanf ("%d",&n);

20         for (i=0; i<n; i++)

21             scanf ("%d %d",&a[i], &b[i]);

22         for (i=0; i<n; i++)//两个for循环排序，没用结构体，比较麻烦

23         {

24             for (j=i; j<n; j++)

25             {

26                 if (a[i] > a[j])

27                 {

28                     swap(a[i], a[j]);

29                     swap(b[i], b[j]);

30                 }

31                 if (a[i] == a[j])

32                 {

33                     if (b[i] > b[j])

34                     {

35                         swap(a[i], a[j]);

36                         swap(b[i], b[j]);

37                     }

38                 }

39 

40             }

41         }

42         time = 0;

43         memset(vis, 0, sizeof(vis));

44         for (i=0; i<n; i++)

45         {

46             if (vis[i])//若该木棍被访问过，跳过

47                 continue;

48             //x = a[i];

49             y = b[i];

50             for (j=i+1; j<n; j++)

51             {

52                 if (!vis[j] && b[j] >= y)

53                 {

54 

55                     vis[j] = 1;//标记访问

56                     //x = a[j];

57                     y = b[j];//当前最大重量

58                 }

59             }

60             time++;//时间加一

61         }

62         printf ("%d\n",time);

63     }

64     return 0;

65 }

View Code