poj 1065 Wooden Sticks（偏序）

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Wooden Sticks

http://poj.org/problem?id=1065

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24062		Accepted: 10369

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 ( 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 ) .

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

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

Source

Taejon 2001

 

题意：

给出n个二元组（a，b），将这n个二元组分成k个部分，每个部分内ai+1>=ai && bi+1>=bi

问最少划分数

 

先将给出的二元组从小到大排序

 

定义（A,≤）是偏序集，≤ 表示 xi+1>=xi && yi+1>=yi

 

那么根据 Dilworth 定理 及其 对偶定理

定理1 令（X,≤）是一个有限偏序集，并令r是其最大链的大小。则X可以被划分成r个但不能再少的反链。 
定理2 令（X,≤）是一个有限偏序集，并令m是反链的最大的大小。则X可以被划分成m个但不能再少的链。

 

推荐一篇写偏序的博客：http://blog.csdn.net/xuzengqiang/article/details/7266034

 

这里最少划分数即为最少链数=最大反链长度

即答案=最长严格递减子序列长度

 

 

#include<cstdio>
#include<iostream>
#include<algorithm>

#define N 5001

using namespace std;

struct node
{
    int a,b;
}e[N];

int s,f[N];

bool cmp(node p,node q)
{
    if(p.a!=q.a) return p.a<q.a;
    return p.b<q.b;
}

int find(int w)
{
    int l=1,r=s,mid,tmp=-1;
    while(l<=r)
    {
        mid=l+r>>1;
        if(f[mid]<=w) tmp=mid,r=mid-1;
        else l=mid+1;
    }
    return tmp;
}

int main()
{
    int T,n,pos;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d%d",&e[i].a,&e[i].b);
        sort(e+1,e+n+1,cmp);
        f[s=0]=2e9;
        for(int i=1;i<=n;i++)
            if(e[i].b<f[s]) f[++s]=e[i].b;
            else 
            {
                pos=find(e[i].b);
                if(pos>0) f[pos]=e[i].b;
            }
        printf("%d\n",s);
    }
}