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。
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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; } |
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