poj1065&1548 dp,最长上升子序列,偏序与反偏序

题目描述：

　　给定n个数对(xi,yi),对这些数对进行划分，保证划分出来的每一堆都是一条单调的链，问最少可以划分成多少堆

　　原题如下：

Wooden Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26409		Accepted: 11423

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

poj1548原理相同，不放了

　　首先我们要求单调链，把样例中的点画在坐标系中，我们便可看出，要求出单调链，要对数据排序（二维降一维）

　　poj讨论区有较为清楚的描述（本人理解但是无法用语言描述）

第一次贪心：其实并非每次都要求一个最长的序列，然后删除，只要选出的序列符合"极大"即可。这里的极大和函数的极值是一个概念。
例如，(1,4) (4,4) (5,3) (9,4) 其中最大(最长)的一个序列为 (1,4)->(4,4)->(9,4)
极大的序列 为 (5,3)->(9,4)
证明：很简单，因为所有的木头都需要被加工，并且与加工的先后顺序无关，那么该木头一定
需要在某个序列中，而我们每次选择的序列是极大的，故每次操作都没有遗漏可加入的木头，因此总的操作数是最少的。
　

　　然后我们就要开始求单调链的数量，这里用到一个定理：（这个定理我不知道啊。。。QAQ）

　　对于Dilworth定理，给出如下解释：

　　一种Dilworth定理的等价表述是：在有穷偏序集中，任何反链最大元素数目等于任何将集合到链的划分中链的最小数目。一个关于无限偏序集的理论指出，在此种情况下，一个偏序集具有有限的宽度w，当且仅当它可以划分为最少w条链。

（摘自https://baike.baidu.com/item/Dilworth%E5%AE%9A%E7%90%86/5489361）

　　可知，对于已经排序好的数据，上升单调链条数=下降单调链的最大长度(任何反链最大元素数目等于任何将集合到链的划分中链的最小数目)

　　所以这道题最终转化成了求最长下降子序列的问题

　　代码如下:

　　

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
    int l,w;
}wood[10000];
bool cmp(node a,node b)
{
    if(a.l==b.l) return a.w<b.w;
    return a.l<b.l;
}
int dp[10000];
int T,n;
void solve()
{
    for(int i=1;i<=n;i++)
    {
        dp[i]=1;
        for(int j=1;j<=i;j++)
        {
            if(wood[j].w>wood[i].w)
                dp[i]=max(dp[i],dp[j]+1);
        }
        //printf("now:%d
",dp[i]);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    cin>>T;
    for(int i=1;i<=T;i++)
    {
        cin>>n;
        for(int j=1;j<=n;j++)
        {
            scanf("%d %d",&wood[j].l,&wood[j].w);
        }
        sort(wood+1,wood+1+n,cmp);
        //for(int j=1;j<=n;j++) printf("%d %d
",wood[j].l,wood[j].w);
        solve();int ans=0;
        for(int j=1;j<=n;j++) ans=max(ans,dp[j]);
        printf("%d
",ans);
    }
    return 0;
}