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 (4,9), (5,2),
(2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is
a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
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 n 2 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
Output for the Sample Input
2
1
3
一开始一直把题目理解错了。。。。以为选一个wooden l w,后面的只要l*>=l&&w*>w,就不用消耗时间
原本题目意思是：先选一个wooden1 l w,紧跟后面的wooden2 l* w*,只要l*>l&&w*>w,wooden2就不要消耗时间，在后面的wooden3要跟wooden2 的 l w 比较，而不是 wooden1,
此题可以先根据 l进行排序，然后用贪心
#include<iostream> #include<algorithm> #include<string.h> #include<stdlib.h> using namespace std; struct as { int l; int w; }p[5010]; int cmp (as a,as b) { if(a.l>b.l) return 0 ; else if(a.l==b.l) { if(a.w>=b.w) return 0; else return 1; } else return 1; } int main() { int t,n,i,j,q[5010],pos; cin >> t; while( t-- ) { memset(q,0,sizeof(q)); cin >>n ; for( i=0;i<n;i++ ) cin>>p[i].l>>p[i].w; sort ( p,p+n,cmp ); for( i=0;i<n;i++ ) if(q[i]==0) { pos=i; for(j=i+1;j<n;j++) if(q[j]==0&&p[j].w>=p[pos].w) { q[j]=1;pos=j; } } int sum=0; for(i=0;i<n;i++) if(q[i]==0) sum++; cout<<sum<<endl; } }