poj3636--Nested Dolls

Nested Dolls

Description

Dilworth is the world‘s most prominent collector of Russian nested dolls: he literally has thousands of them! You know, the wooden hollow dolls of different sizes of which the smallest doll is contained in the second smallest, and this doll is in turn contained in the next one and so forth. One day he wonders if there is another way of nesting them so he will end up with fewer nested dolls? After all, that would make his collection even more magnificent! He unpacks each nested doll and measures the width and height of each contained doll. A doll with width w₁ and height h₁ will fit in another doll of width w₂ and height h= if and only if w₁ < w₂ and h₁ < h₂. Can you help him calculate the smallest number of nested dolls possible to assemble from his massive list of measurements?

Input

On the first line of input is a single positive integer 1 ≤ t ≤ 20 specifying the number of test cases to follow. Each test case begins with a positive integer 1 ≤ m ≤ 20000 on a line of itself telling the number of dolls in the test case. Next follow 2m positive integers w₁, h₁,w₂, h₂, ... ,w_m, h_m, where w_i is the width and h_i is the height of doll number i. 1 ≤ w_i, h_i ≤ 10000 for all i.

Output

For each test case there should be one line of output containing the minimum number of nested dolls possible.

Sample Input

4
3
20 30 40 50 30 40
4
20 30 10 10 30 20 40 50
3
10 30 20 20 30 10
4
10 10 20 30 40 50 39 51

Sample Output

1
2
3
2

Source

Nordic 2007

//A doll with width w₁ and height h₁ will fit in another doll of width w₂ and height h= if and only if w₁ < w₂ and h₁ < h_{2  有没有等号会影响bool  函数中的返回值 ；l相同的w要按照降序排列，l升序排列;}

 1 #include <stdio.h>
 2 #include <algorithm>
 3 using namespace std ;
 4
 5 struct dool
 6 {
 7     int l;
 8     int w;
 9 } ;
10 dool num[20020] ;
11
12 bool cmp(dool l, dool w)  //Dirworth定理：l按照升序排序；当l相等时，w也降序，排除了等号关系；
13 {
14     if(l.l == w.l)
15     return l.w > w.w ;
16     else
17     return l.l < w.l ;
18 }
19
20 int main()
21 {
22     int m,n,i ;
23     scanf("%d",&m) ;
24     while(m--)
25     {
26         scanf("%d",&n) ;
27         for(i=0; i<n; i++)
28         scanf("%d %d",&num[i].l, &num[i].w) ;
29         sort(num, num+n, cmp) ;
30
31         int temp, total=0, j ;
32         for(i=0; i<n; i++)
33         {
34             if(num[i].w != 0)
35             {
36                 temp=num[i].w ;
37                 total++ ;
38                 for(j=i+1; j<n; j++)
39                 {
40                     if(num[j].w > temp)
41                     {
42                         temp = num[j].w ;
43                         num[j].w = 0 ;
44                     }
45                 }
46             }
47         }
48         printf("%d\n",total) ;
49     }
50     return 0;
51 }