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。

 1    #include <iostream>
 2    #include <cstdio>
 3    #include <cstring>
 4    #include <cmath>
 5    #include <algorithm>
 6    #include <iomanip>
 7    
 8    
 9    using namespace std;
10    
11    int main()
12    {
13        int t;
14        cin>>t;
15        int n;
16        int l[6666],w[6666],lk[6666],wk[6666];
17        while (t--)
18        {
19            memset(l,0,sizeof(l));
20            memset(w,0,sizeof(w));
21            memset(lk,0,sizeof(lk));
22            memset(wk,0,sizeof(wk));
23            scanf("%d",&n);
24            for (int i=1;i<=n;i++)
25                scanf("%d %d",&l[i],&w[i]);
26            for (int i=1;i<=n-1;i++)
27                for (int j=i+1;j<=n;j++)
28                 if ((l[i]>l[j])||((l[i]==l[j])&&(w[i]>w[j])))
29            {
30                swap(l[i],l[j]);
31                swap(w[i],w[j]);
32    
33            }
34    
35                int k;
36                int sum=1;
37                lk[1]=l[1];
38                wk[1]=w[1];
39              //  for (int i=1;i<=n;i++)
40            //        cout<<"look"<<l[i]<<"  "<<w[i]<<endl;
41             for (int i=1;i<=n;i++)
42             {
43    
44                 k=1;
45                 while ((l[i]<lk[k])||(w[i]<wk[k]))
46                 {
47                     k++;
48                 //    cout<<"look"<<endl;
49                     if (k>sum) break;
50                 }
51                 if (k>sum)
52                 {
53                     sum++;
54                     lk[k]=l[i];
55                     wk[k]=w[i];
56                 }
57                 else
58                 {
59                     lk[k]=l[i];
60                     wk[k]=w[i];
61                 }
62             //    cout<<"k:"<<k<<endl;
63    
64    
65             }
66                  printf("%d\n",sum);
67        }
68        return 0;
69    }
70
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