POJ-1065 Wooden Sticks
POJ-1065 Wooden Sticks
题面
Problem 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.
题意
英文看半天看不懂？
翻译：有N根木棍等待处理。机器在处理第一根木棍时需要准备1分钟，此后遇到长宽都不大于第一根木棍的木棍就不需要时间准备，反之则需要1分钟重新准备。比如木棍按照(3,3)、(1,3)、(1,4)、(2,3)的顺序进入，共需要准备3分钟
还是翻译好QWQ。。
我的想法是，先排序，然后对于完全小于的东西，我们可以丢掉他，剩下的那些无法比较的就是答案。（待证明）
实际上就是给出一些偏序关系，求chain的最小值。
Dilworth定理：最少的chain个数等于最大的antichain的大小。
相当于求按l排序后，w的最长下降子序列。
代码
#include <iostream>
#include <algorithm>
#include <cstring>
//#include <bits/stdc++.h>
using namespace std;
int T,n;
pair<int,int> p[5010];
int a[5010];
int INF=0x7ffffff7;
int cmp(pair<int,int> q,pair<int,int> w)
{
return q<w;
}
int main()
{
cin>>T;
while (T--)
{
cin>>n;
for (int i=1;i<=n;i++) cin>>p[i].first>>p[i].second;
sort(p+1,p+n+1,cmp);
for (int i=1;i<=n+1;i++) a[i]=INF;
for (int i=1;i<=n;i++) *lower_bound(a+1,a+n+2,-p[i].second)=-p[i].second;
cout<<distance(a+1,lower_bound(a+1,a+n+2,INF))<<endl;
}
}
