1051 Wooden Sticks（贪心-3）

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

 1 #include<iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <stdio.h>
 5 #include <math.h>
 6 using namespace std;
 7 struct sa
 8 {
 9     int x;
10     int y;
11     int flag;
12 }data[10005];
13 int cmp(const sa &a,const sa &b)
14 {
15     if(a.x!=b.x)
16     return a.x<b.x;
17     else
18     return a.y<b.y;
19 }
20 int main()
21 {
22     int n,m,ans,tmp1,tmp2;
23     while(cin>>n)
24     {
25         while(n--)
26         {
27             cin>>m;
28             for(int i=0;i<m;i++)
29             {
30                 cin>>data[i].x>>data[i].y;
31                 data[i].flag=0;
32             }
33             sort(data,data+m,cmp);
34             ans=0;
35             for(int i=0;i<m;i++)
36             {
37                 if(data[i].flag!=1)
38                 {
39                     ans++;
40                     tmp1=data[i].x;tmp2=data[i].y;data[i].flag=1;
41                     for(int j=i+1;j<m;j++)
42                     if(data[j].x>=tmp1&&data[j].y>=tmp2&&data[j].flag==0)
43                     {
44                         tmp1=data[j].x;tmp2=data[j].y;data[j].flag=1;
45                     }
46                 }
47             }
48             cout<<ans<<endl;
49         }
50     }
51     return 0;
52 }

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