POJ 1065 Wooden Sticks （贪心）

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

贪心基础上加个id递增的条件

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<vector>
 5 #include<cstring>
 6 #include<string>
 7 #include<algorithm>
 8 #include<map>
 9 #include<cmath>
10 #include<math.h>
11 using namespace std;
12 
13 struct node
14 {
15     int l,w;
16 }a[5005];
17 
18 bool vis[5005];
19 
20 bool cmp(node b,node c)
21 {
22     if(b.l==c.l)
23         return b.w<c.w;
24     else
25         return b.l<c.l;
26 }
27 
28 int main()
29 {
30     
31     int T;
32     scanf("%d",&T);
33     while(T--)
34     {
35         int n;
36         scanf("%d",&n);
37         for(int i=0;i<n;i++)
38             scanf("%d%d",&a[i].w,&a[i].l);
39         sort(a,a+n,cmp);
40         int res=0;
41         memset(vis,false,sizeof(vis));
42         int tmpl,tmpw;
43         for(int i=0;i<n;i++)
44         {
45             if(!vis[i])
46             {
47                 vis[i]=true;
48                 res++;
49                 tmpl=a[i].l;
50                 tmpw=a[i].w;
51                 for(int j=i+1;j<n;j++)
52                 {
53                     if(a[j].l>=tmpl&&a[j].w>=tmpw&&!vis[j])
54                     {
55                         vis[j]=true;
56                         tmpl=a[j].l;
57                         tmpw=a[j].w;
58                     }
59                 }
60             }
61         }
62         printf("%d
",res);
63     }
64     return 0;
65 }