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，我很注意节制的（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 次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。
代码实现
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 }