poj 1065 Wooden Sticks

Wooden Sticks
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 19008 Accepted: 8012
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

 

 

 

贪心题。题目很容易懂，就不翻译了。略水，一遍AC，久违的快感！！！hhhh ，嘛，我为注意节制的（哪和哪！）

把w和l分别按第一关键字和第二关键字排序。

然后扫描一遍，如果符合w[i]>=wk[k]&&l[i]>=lk[k]就更新wk[k]和lk[k]，分别表示的是第K的操作下能达到的最大w和l.

如果不符合，则需要增加一分钟的工作时间。同样不要忘记更新。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <iomanip>

 7 

 8 

 9 using namespace std;

10 

11 int main()

12 {

13     int t;

14     cin>>t;

15     int n;

16     int l[6666],w[6666],lk[6666],wk[6666];

17     while (t--)

18     {

19         memset(l,0,sizeof(l));

20         memset(w,0,sizeof(w));

21         memset(lk,0,sizeof(lk));

22         memset(wk,0,sizeof(wk));

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

24         for (int i=1;i<=n;i++)

25             scanf("%d %d",&l[i],&w[i]);

26         for (int i=1;i<=n-1;i++)

27             for (int j=i+1;j<=n;j++)

28              if ((l[i]>l[j])||((l[i]==l[j])&&(w[i]>w[j])))

29         {

30             swap(l[i],l[j]);

31             swap(w[i],w[j]);

32 

33         }

34 

35             int k;

36             int sum=1;

37             lk[1]=l[1];

38             wk[1]=w[1];

39           //  for (int i=1;i<=n;i++)

40         //        cout<<"look"<<l[i]<<"  "<<w[i]<<endl;

41          for (int i=1;i<=n;i++)

42          {

43 

44              k=1;

45              while ((l[i]<lk[k])||(w[i]<wk[k]))

46              {

47                  k++;

48              //    cout<<"look"<<endl;

49                  if (k>sum) break;

50              }

51              if (k>sum)

52              {

53                  sum++;

54                  lk[k]=l[i];

55                  wk[k]=w[i];

56              }

57              else

58              {

59                  lk[k]=l[i];

60                  wk[k]=w[i];

61              }

62          //    cout<<"k:"<<k<<endl;

63 

64 

65          }

66               printf("%d\n",sum);

67     }

68     return 0;

69 }