HDU-1677

Nested Dolls

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3796 Accepted Submission(s): 1161

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

题意是套娃，每个套娃有自己的宽度和高度只有w、h都严格小于当前套娃才可以嵌套，问最少可以剩几个套娃。

先对w做从大到小排序，对h做从小到大排序。为了尽可能多的嵌套

将所有套娃尺寸标记为INF，每当出现一个不可套就将套娃的尺寸转移到标记。

最后非INF的个数就是套娃的个数。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int INF=10010;
 5 
 6 struct doll{
 7     int w,h;
 8 }d[20010],dp[20010];
 9 
10 bool cmp(doll a,doll b){
11     if(a.w==b.w)
12     return a.h<b.h;
13     else
14     return a.w>b.w;
15 }
16 
17 int main(){
18     int t,n;
19     cin>>t;
20     while(t--){
21         cin>>n;
22         for(int i=0;i<n;i++){
23             cin>>d[i].w>>d[i].h;
24             dp[i].w=INF;
25             dp[i].h=INF;
26         }
27         sort(d,d+n,cmp);
28         //cout<<d[0].w<<endl;
29         for(int i=0;i<n;i++){
30             int j=0;
31             while(dp[j].w<=d[i].w||dp[j].h<=d[i].h)
32             j++;
33             dp[j].w=d[i].w;
34             dp[j].h=d[i].h;
35         }
36         int ans=0;
37         for(int i=0;i<n;i++){
38             if(dp[i].h!=INF)
39             ans++;
40         }
41         cout<<ans<<endl;
42     }
43     return 0;
44 }