Wooden Sticks
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 19 Accepted Submission(s) : 7
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Problem 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 (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
Sample Output
2
1
3
Source Asia 2001, Taejon (South Korea) AC代码：
#include
#include
#include
#include
using namespace std; struct wooden{ int l,w; }; wooden my[5010]; bool comp(wooden a,wooden b){ if(a.l>b.l)return 1; else if(a.l==b.l) return a.w>b.w; else return 0; } int main(){ int t; cin>>t; while(t--){ int n; cin>>n; int i=0,j; while(i
>my[i++].l>>my[i].w; sort(my,my+n,comp); //for(i=0;i
=my[i].l&&my[j].w>=my[i].w){ out--; my[j].l=my[i].l; my[j].w=my[i].w; my[i].l=0; my[i].w=0; break; } } // for(i=0;i
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