hd1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15221 Accepted Submission(s): 6246

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
   
   
   
   

思路：先按木条长度递减排序，若相等按重量大的优先，然后遍历整个数据，符合条件就用一个平行数组标记遍历过，遍历完一组计数1，循环整个即可！

我的这种其实就是最笨的遍历！凑乎的过了！

#include <iostream>
#include<algorithm>
using namespace std;
typedef struct{
int l,w;}sticks;

bool cmp(sticks a,sticks b)
{
    if(a.l == b.l)
     return a.w > b.w;
    else
     return a.l > b.l;
}

int main()
{
    sticks a[5001];
    int t,n,i,count,record;
    cin >>t;
    while(t--)
    {
        int sign[5001]={0};//用于标记木条是否遍历过
        count = 0;//记录时间
        cin >> n;
        for(i=0;i<n;i++)
          cin >> a[i].l >> a[i].w;
        sort(a,a+n,cmp);//按长度递减排序，如果相等重量大的优先
       /* for(i=0;i<n;i++)
         cout << a[i].l <<" "<< a[i].w << endl;*/
      
          for(i=0;i<n;i++)
          {
              record = i;//记录起始木条位置
              if(sign[record] == 0)//判断此木条是否遍历过
              {
                  for(int j=record+1;j<n;j++)
                  {
                    if(a[j].w <= a[record].w &&sign[j] == 0)
                    {
                      record = j;
                      sign[j] = 1;//标记此木条已遍历
                    }
                  }
                  count++;//时间增加一

              }
          }
         cout << count << endl;
    }

    return 0;
}