ZOJ - 1025 Wooden Sticks

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

Output for the Sample Input

2 
1 
3

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 5000 + 10;
const int maxw = 10000 + 10;
struct Node { int x, y, vis; } stk[maxn];
bool cmp(const Node& a, const Node& b)
{
    return a.x > b.x;
}

int main()
{
	int T, n;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		for(int i=0; i<n; i++) {
			scanf("%d%d", &stk[i].x, &stk[i].y);
			stk[i].vis = 0;
		}
		sort(stk, stk+n, cmp);
		int yes = 1, cnt = 0;
		while(yes) {
			int mx = maxw;
			yes = 0; cnt++;
			for(int i=0; i<n; i++) {
				if(mx >= stk[i].y && !stk[i].vis) {
					yes = 1;
					mx = stk[i].y;
					stk[i].vis = 1;
				}
			}
		}
		printf("%d\n", cnt-1);
	}
    return 0;
}