POJ 1065 Wooden Sticks （贪心）

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

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 2n 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

贪心基础上加个id递增的条件

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<vector>
 #include<cstring>
 #include<string>
 #include<algorithm>
 #include<map>
 #include<cmath>
 #include<math.h>
 using namespace std;

 struct node
 {
     int l,w;
 }a[];

 bool vis[];

 bool cmp(node b,node c)
 {
     if(b.l==c.l)
         return b.w<c.w;
     else
         return b.l<c.l;
 }

 int main()
 {

     int T;
     scanf("%d",&T);
     while(T--)
     {
         int n;
         scanf("%d",&n);
         for(int i=;i<n;i++)
             scanf("%d%d",&a[i].w,&a[i].l);
         sort(a,a+n,cmp);
         int res=;
         memset(vis,false,sizeof(vis));
         int tmpl,tmpw;
         for(int i=;i<n;i++)
         {
             if(!vis[i])
             {
                 vis[i]=true;
                 res++;
                 tmpl=a[i].l;
                 tmpw=a[i].w;
                 for(int j=i+;j<n;j++)
                 {
                     if(a[j].l>=tmpl&&a[j].w>=tmpw&&!vis[j])
                     {
                         vis[j]=true;
                         tmpl=a[j].l;
                         tmpw=a[j].w;
                     }
                 }
             }
         }
         printf("%d\n",res);
     }
     return ;
 }