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 ",sum); 67 } 68 return 0; 69 }