poj 1065 Wooden Sticks

Wooden Sticks
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 19008 Accepted: 8012
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 ( 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

 

 

 

贪心题。题目很容易懂，就不翻译了。略水，一遍AC，久违的快感！！！hhhh ，嘛，我为注意节制的（哪和哪！）

把w和l分别按第一关键字和第二关键字排序。

然后扫描一遍，如果符合w[i]>=wk[k]&&l[i]>=lk[k]就更新wk[k]和lk[k]，分别表示的是第K的操作下能达到的最大w和l.

如果不符合，则需要增加一分钟的工作时间。同样不要忘记更新。

C++
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iomanip>
 
 
using namespace std;
 
int main()
{
    int t;
    cin>>t;
    int n;
    int l[6666],w[6666],lk[6666],wk[6666];
    while (t--)
    {
        memset(l,0,sizeof(l));
        memset(w,0,sizeof(w));
        memset(lk,0,sizeof(lk));
        memset(wk,0,sizeof(wk));
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            scanf("%d %d",&l[i],&w[i]);
        for (int i=1;i<=n-1;i++)
            for (int j=i+1;j<=n;j++)
             if ((l[i]>l[j])||((l[i]==l[j])&&(w[i]>w[j])))
        {
            swap(l[i],l[j]);
            swap(w[i],w[j]);
 
        }
 
            int k;
            int sum=1;
            lk[1]=l[1];
            wk[1]=w[1];
          //  for (int i=1;i<=n;i++)
        //        cout<<"look"<<l[i]<<"  "<<w[i]<<endl;
         for (int i=1;i<=n;i++)
         {
 
             k=1;
             while ((l[i]<lk[k])||(w[i]<wk[k]))
             {
                 k++;
             //    cout<<"look"<<endl;
                 if (k>sum) break;
             }
             if (k>sum)
             {
                 sum++;
                 lk[k]=l[i];
                 wk[k]=w[i];
             }
             else
             {
                 lk[k]=l[i];
                 wk[k]=w[i];
             }
         //    cout<<"k:"<<k<<endl;
 
 
         }
              printf("%d\n",sum);
    }
    return 0;
}
 
 
 

作者： CrazyKK

ex-ACMer@hust，researcher@sensetime

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