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;
}