Wooden Sticks

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

题解:

这道题目是先要排序的,按照长度或者重量排都可以,当长度(重量)相同时就按照重量(长度)排,从大到小或从小到大都可以!这里我懂的,没有问题!
排序之后,问题就可以简化,(假设按照长度不等时长度排,长度等是按照重量排,我假设按照从大到小来排!)即求排序后的所有的重量值最少能表示成几个集合。长度就不用再管了,从数组第一个数开始遍历,只要重量值满足条件,那么这两个木棍就满足条件!
 

 #include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
struct wooden{
int l,w;
};
wooden my[];
bool comp(wooden a,wooden b){
if(a.l>b.l)return ;
else if(a.l==b.l)
return a.w>b.w;
else return ;
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
int i=,j;
while(i<n)
{
scanf("%d %d",&my[++i].l,&my[i].w);
}
sort(my,my+n,comp);
int out=n;
for(i=;i<n;i++)
for(j=;j<=i-;j++){
if(my[j].l>=my[i].l&&my[j].w>=my[i].w){
out--;
my[j].l=my[i].l;
my[j].w=my[i].w;
my[i].l=;
my[i].w=;
break;
}
}
printf("%d\n",out);
}
return ;
}

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