Wooden Sticks

time limit per test 1 second memory limit per test 256 megabytes
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 ) .

传送门:POJ1065

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

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2
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4
5
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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

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3
2
1
3

题解

dilworth定理的运用 对于一个偏序集 如果把它划分成链 最少的链的个数等于最大的反链长度
http://www.cnblogs.com/fstang/archive/2013/03/31/2991255.html 这篇博客讲的非常好
那么这道题是一个二维的偏序集 只要 先对l从小到大排序 l相同时w小的在前面(避免计算反链时多计算) 再计算一下w的最长递减子序列就可以了

AC code:(不包含输入类)

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import java.io.*;  
import java.util.*;
public class Main {

public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter pw = new PrintWriter(System.out);
int t=sc.nextInt();
while (sc.hasNext()) {
int n=sc.nextInt();
Node[]node=new Node[n];
for(int i=0;i<n;i++){
node[i]=new Node(sc.nextInt(),sc.nextInt());
}
Arrays.sort(node);
ArrayList<Integer>list=new ArrayList<Integer>();
for(int i=0;i<n;i++){
if(list.size()==0){
list.add(node[i].r);
}else{
if(node[i].r<list.get(list.size()-1)){
list.add(node[i].r);
}else{
int index=find(node[i].r,list);
list.set(index, node[i].r);
}
}
}
int l=list.size();
pw.println(l);
pw.flush();
}
}
static int find(int k,ArrayList<Integer>list){
int start=0;
int end=list.size()-1;
int result=0;
while(start<=end){
int mid=(start+end)/2;
if(list.get(mid)<=k){
result=mid;
end=mid-1;
}else{
start=mid+1;
}
}
return result;
}
}
class Node implements Comparable<Node>{
int l;
int r;
Node(int a,int b){
l=a;
r=b;
}
@Override
public int compareTo(Node o) {
if(this.l<o.l)return -1;
if(this.l>o.l)return 1;
if(this.r<o.r)return -1;
if(this.r>o.r)return 1;
return 0;
}
}