Problem descriptionthere 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
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 (), and ), then the minimum setup time shocould be 2 minutes since there is
A sequence of pairs ).
Inputthe 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 magnloud 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.
Outputthe output shoshould 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
213
#include <iostream>#include <algorithm>using namespace std;struct Node{ int head; int end; int flag;} node[5005];int cmp(struct Node a,struct Node b){ if(a.head < b.head) return 1; else if(a.head == b.head) return a.end < b.end; return 0;}int main(){ int n,t; cin >> t; while(t--) { cin >> n; int i,j,sum = 0,cnt = 0; for(i = 0; i<n; i++) { cin >> node[i].head >> node[i].end; node[i].flag = 0; } sort(node,node+n,cmp); int ans = 0; for(i = 0; i<n; i++) { if(!node[i].flag) { node[i].flag = 1; ans++; int end = node[i].end; for(int j = i+1; j < n; j++) { if(!node[j].flag && node[j].end >= end) { node[j].flag = 1; end = node[j].end; } } } } cout << ans << endl; } return 0;}