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

213

 

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 IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor153260762015-10-31 23:25:00Accepted105115MS1808K1037 BC++BossJue