[置顶] hdu 1051 Wooden Sticks(Asia 2001, Taejon (South Korea))

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 (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

 

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 n 2 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

Asia 2001, Taejon (South Korea)

 

算法：

        还是一个集合划分问题，显然，两个矩形的包含关系为偏序，因而最小的划分数目即是反链集合的最多元素数目，一个集合为反链，即这个集合中任何两个矩形A、B，A不包含B，B也不包含A，如果把一个反链集合中的矩形按照width递增排序，则它们的iheight一定是递减的（否则就会有一个矩形的width和height均小于另一个矩形，这样，它就被另一个矩形所包含，与反链定义矛盾），所以，我们先按照iwidth排序，于是问题转化为iheight的最长递减子序列，复杂度为：O(nlogn)--排序 + O(nlogn)--最长递减子序列 ==  O(nlogn)。

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef struct Record{
public:
	int iWidth;
	int iHeight;
}Record;

Record r[50000];
int n;
const int INF = 0x0FFFFFFF;
int f[50000],g[50000];

void Handle();
int cmp(const void* lhs, const void* rhs);

int main(){
	int icase;
	scanf("%d", &icase);
	while(icase--){
		Handle();
	}
}

void Handle(){
	int idx;
	scanf("%d",&n);
	for(int i = 0; i < n; ++i){
		scanf("%d %d", &r[i].iWidth, &r[i].iHeight);
	}
	qsort(r, n, sizeof(Record), cmp);
	fill(g, g + n, INF);
	for(int i = 0; i < n; ++i){
		idx    = lower_bound(g, g + n, r[i].iHeight) - g;
		f[i]   = idx + 1;
		g[idx] = r[i].iHeight;
	}
	printf("%d\n", *max_element(f, f+ n));
}

int cmp(const void* lhs, const void* rhs){
	const Record* l = (const Record*)lhs;
	const Record* r = (const Record*)rhs;
	if(l->iWidth != r->iWidth){
		return r->iWidth - l->iWidth;
	}else{
		return r->iHeight - l->iHeight;
	}
}

运行结果：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
15326076	2015-10-31 23:25:00	Accepted	1051	15MS	1808K	1037 B	C++	BossJue