/*Problem D
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 14 Accepted Submission(s) : 8
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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
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct p
{
int x, y;
}a[10010];
int cmp(const void *a, const void*b)
{
struct p *c = (struct p *)a;
struct p *d = (struct p *)b;
if( c->x != d->x )
return c->x - d->x;
else
return c->y - d->y;
}
int mark[10010];
int main()
{
int i, j , k, m , n, sum, l, w;
scanf("%d", &n);
while(n--)
{
sum = 0;
scanf("%d", &m);
for(i = 0; i < m; i++)
scanf("%d%d", &a[i].x, &a[i].y);
memset(mark, 0, sizeof(mark));
qsort(a, m, sizeof(a[0]), cmp);
for(i = 0; i < m; i++)
{
if(!mark[i])
{
k = i;
sum++;
}
else
continue;
l = a[i].x;
w = a[i].y;
for(j = k+1; j < m; j++)
{
if(a[j].x >= l && a[j].y >= w && !mark[j])
{
mark[j] = 1;
l = a[j].x;
w = a[j].y;
}
}
}
printf("%d\n", sum);
}
return 0;
}
