| Wooden Sticks |
| Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB |
| 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 |
/// 很简单的一道题，思路先一 length 或者 weight 排序 sticks, 然后以另一个变量(length or weight) 以“ 最大”消逆序，如果不能消则 times 增 １，并将该变量插入链表中（表里面保存的就是逆序的东西） 以
5
4 9 5 2 2 1 3 5 1 4
为例：按 length 排序后为(1,4),(2,1),(3,5),(4,9),(5,2) 首先 4 进入链表 times=1, 下一个 1,因为链表里没有小于等于 １的数所以 1,插入表头，times++. 接下来为, 5,9, 每次我们用表中较大的数来比较，所以链表的变化为 1,5, ==== > 1,9 , 最后 一个 2， 因为表里有1，1<=2成立所以最后一根也不需要发时间，链表变为 2,9 , times=2.
//// MY CODE ， NOTE HOW TO USE STL
#include <iostream>
#include <list>
#include <algorithm>
using namespace std;
int const MAX = 5000;
pair<int,int> stick[MAX];
struct myCmp{
bool operator()(const pair<int,int>& a,const pair<int,int>& b){
return a.first<b.first;
}
};
int main(){
int nTest;
cin>>nTest;
for(int i=0; i<nTest; i++){
int nStks,j;
cin>>nStks;
for(j=0; j<nStks; j++)
cin>>stick[j].first>>stick[j].second;
sort(&stick[0],&stick[j],myCmp());
list<int> lst;
list<int>::reverse_iterator it;
int times=1;
lst.push_back(stick[0].second);
for(int j=1; j<nStks; j++){
for(it=lst.rbegin(); it!=lst.rend(); it++){
if(stick[j].second>=*it){
*it = stick[j].second;
break;
}
}
if(it==lst.rend()){
lst.push_front(stick[j].second);
times++;
}
}
cout<<times<<endl;
}
return 0;
}
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