Nested Dolls

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2704    Accepted Submission(s): 802

Problem 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 h2 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, . . . ,wm, hm, where wi is the width and hi is the height of doll number i. 1 <= wi, hi <= 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

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

题意：  给你N个木偶娃娃，每一个娃娃都有一定的高度，和宽度，然后娃娃的内部是空心的，现在要你把小的娃娃放到大娃娃的肚子去，然后问最后剩下几个娃娃..

  有点像盒子套盒子意思，处理方法。采用DP

LIS二维处理：  现将一个属性比如 height作为标准，从大到小排序，遇到相等的width，则对另一个属性h，从小到大排序。

代码：

 1 //#define LOCAL
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<iostream>
 6 using namespace std;
 7 const int maxn=20005;
 8 const int inf=0x3f3f3f3f;
 9 int m;
10 
11 struct doll
12 {
13    int w,h;
14   bool operator <(const doll &a)const
15   {
16         if(w==a.w)
17          return  h <a.h;    //ÉýÐò
18       else
19          return w > a.w ;          //½µÐò
20   }
21 };
22 
23 
24 doll aa[maxn],ans[maxn];
25 int dp[maxn];
26 
27 int binary(doll v)
28 {
29     int l=1,r=m,mid;
30     while(l<=r)
31     {
32       mid=l+((r-l)>>1);    //降序
33      if(ans[mid].h<=v.h)
34             l=mid+1;
35       else
36           r=mid-1;
37     }
38   return l;
39 }
40 
41 int LIS(doll a[],int n)
42 {
43  int i;
44  int res=0;
45  for(i=1;i<=n;i++)
46  {
47      ans[i].h=inf;
48      ans[i].w=inf;
49  }
50  for(i=1 ; i<=n ; i++){
51      dp[i]=binary(a[i]);
52      if(res<dp[i])res=dp[i];
53      if(ans[dp[i]].h>a[i].h&&ans[dp[i]].w>a[i].w){
54       ans[dp[i]].h=a[i].h;
55       ans[dp[i]].w=a[i].w;
56     }
57  }
58  return res;
59 }
60 
61 int main()
62 {
63   #ifdef LOCAL
64    freopen("test.in","r",stdin);
65   #endif
66 
67     int cas;
68     scanf("%d",&cas);
69    while(cas--){
70      scanf("%d",&m);
71     for(int i=1;i<=m;i++){
72       scanf("%d%d",&aa[i].w,&aa[i].h);
73     }
74     sort(aa+1,aa+m+1);
75     printf("%dn",LIS(aa,m));
76   }
77   return 0;
78 }