HDU1051Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21183    Accepted Submission(s): 8546

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

 

题意：给你n个棍子，和他们的重量W高度h求最小准备时间。第一根木棍的准备时间为1分钟。往后如果W<=W'，H<=H'的准备时间为0分钟，反之为1分钟。

思路：先按照W对他们进行排序，以后按照H从上往下找比他大的

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
    int x,y;
}a[5001];
int vis[5001];
int cmp(node b,node c )
{
    return b.x<=c.x;
}
int main()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        int n;
        cin>>n;
        for(int j=0;j<n;j++)
        {
            cin>>a[j].x>>a[j].y;
        }
        sort(a,a+n,cmp);
       int s=0;
       memset(vis,0,sizeof(vis));
       for(int j=0;j<n;j++)
       {
           int p=j;
           if(!vis[j])
           {
               vis[j]=1;
                for(int k=j+1;k<n; k++)
               {
                   if(!vis[k]&&a[p].y<=a[k].y)
                   {
                       vis[k]=1;
                        p=k;
                      continue;


                   }
               }
               s++;
           }


       }
       cout<<s<<endl;
    }
    return 0;
}