poj 1065 Wooden Sticks

Wooden SticksTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19008 Accepted: 8012Description
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
213
 
 
 
탐욕 문제.제목을 알기 쉬워서 번역하지 않겠다.약수, AC 한 번, 오랜만에 쾌감!!!hhh, 글쎄, 내가 주의하기 위해 절제한 (어디와!)
w와 l를 각각 첫 번째 키워드와 두 번째 키워드로 정렬합니다.
그리고 한 번 스캔해서 w[i]>=wk[k] & l[i]>=lk[k]에 맞으면 wk[k]와 lk[k]를 업데이트합니다. 각각 K의 조작에서 달성할 수 있는 최대 w와 l를 표시합니다.
만약 부합되지 않는다면 1분의 작업 시간을 늘려야 한다.마찬가지로 업데이트를 잊지 마세요.

 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
",sum);

67     }

68     return 0;

69 }