Wooden Sticks -HZNU寒假集训

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 ;
 }