11345 - Rectangles

[shakil]

Why WA.........

Code: Select all

Cut after AC.....

[rio]

Try this.
INPUT:

Code: Select all

1
2
0 0 1 1
1 1 0 0

OUTPUT:

Code: Select all

Case 1: 0

ADD : If you meet these kind of problems, try everything. Finding the trick is also the part of the problem.
----
Rio

[saman_saadi]

Try this.
INPUT:

Code: Select all

1
2
0 0 1 1
1 1 0 0

OUTPUT:

Code: Select all

Case 1: 0

ADD : If you meet these kind of problems, try everything. Finding the trick is also the part of the problem.
----
Rio

Are you sure? I think the answer is 1 not 0. because two rectangles are the same.

[saman_saadi]

Is there any tricky IO?

[sclo]

Is there any tricky IO?

All I did was to find the maximum x1, minimum x2, maximum y1 and minimum y2 over all input.
As a side effect, if x2<=x1 or y2<=y1 in any rectangle, the answer is 0

[saman_saadi]

Is there any tricky IO?

All I did was to find the maximum x1, minimum x2, maximum y1 and minimum y2 over all input.
As a side effect, if x2<=x1 or y2<=y1 in any rectangle, the answer is 0

Thanks sclo for your quick reply. Your method is much better than mine and it works correctly but I don't know why my method is wrong. My algorithms is:

I use a priority queue that the rectangle with smallest x1 exit first. I push all input rectangles into priority queue. Suppose r1 and r2 be the first and the second rectangle that we pop from the priority queue:
the answer is 0 if (r2.x1 >= r1.x2 or (r2.y1 >= r1.y2 or r2.y2 <= r1.y1)) otherwise we make a new rectangle that we call r:
r.x1 = r2.x1
r.y1 = max(r1.y1, r2.y1)
r.x2 = min(r1.x2, r2.x2)
r.y2 = min(r1.y2, r2.y2)
then we push r into priority queue.we repeat above algorithm until the priority queue size is equal to one or we found the answer is 0.

[sclo]

what happens when r.x1>r.x2 or r.y1>r.y2 in your algorithm?

[saman_saadi]

what happens when r.x1>r.x2 or r.y1>r.y2 in your algorithm?

I think I handle all of these:
1) r.x1 > r.x2 => r2.x1 > min(r1.x2, r2.x2)
we know r2.x1 < r2.x2 so we must have r1.x2 < r2.x1 that I handle this.
2) r.y1 > r.y2 => max(r1.y1, r2.y1) > min(r1.y2, r2.y2)
suppose that min(r1.y2, r2.y2) = r1.y2:
r1.y2 < max(r1.y1, r2.y1) => r1.y2 < r2.y1 that I handle this
now suppose min(r1.y2, r2.y2) = r2.y2:
r2.y2 < max(r1.y1, r2.y1) => r2.y2 < r1.y1 that I handle this.
So I handle all cases that you consider.

[sclo]

what happens when r.x1>r.x2 or r.y1>r.y2 in your algorithm?

I think I handle all of these:
1) r.x1 > r.x2 => r2.x1 > min(r1.x2, r2.x2)
we know r2.x1 < r2.x2 so we must have r1.x2 < r2.x1 that I handle this.
2) r.y1 > r.y2 => max(r1.y1, r2.y1) > min(r1.y2, r2.y2)
suppose that min(r1.y2, r2.y2) = r1.y2:
r1.y2 < max(r1.y1, r2.y1) => r1.y2 < r2.y1 that I handle this
now suppose min(r1.y2, r2.y2) = r2.y2:
r2.y2 < max(r1.y1, r2.y1) => r2.y2 < r1.y1 that I handle this.
So I handle all cases that you consider.

Obviously, something is wrong with your logic somewhere. But I don't really see it.

[Masud_CSE_SUST]

what's wrong? why WA?

Code: Select all

code removed

any more I/O

[shakil]

if(t.x2>=t.x1&&t.y2>=t.y1)
area = ( t.x2 - t.x1 ) * ( t.y2 - t.y1 ) ;
else
area = 0 ;
hope it help.
and cut this->

if( r.x2 <= t.x1 || t.x2 <= r.x1 || r.y2 <= t.y1 || t.y2 <= r.y1 )
{
area = 0 ; break ;
}

[Masud_CSE_SUST]

again WA

[rio]

again WA

How did you fixed your code ?
Update your code, or erase it and post a new one.

-----
Rio

[Masud_CSE_SUST]

rio wrote

How did you fixed your code ?
Update your code, or erase it and post a new one.

Updated code (got WA)

Code: Select all

#include <stdio.h>

struct ractangle
{
	int x1, y1, x2, y2 ;
} ;

int main ()
{
	//freopen( "11345.in", "r", stdin ) ;

	int T, kase, i, N, area ;
	ractangle r[50], t ;

	scanf( "%d", &T ) ;

	kase = 1 ;

	while( T-- )
	{
		scanf( "%d", &N ) ;

		for( i = 1; i<= N; i++ )
		{
			scanf( "%d%d%d%d", &r[i].x1, &r[i].y1, &r[i].x2, &r[i].y2 ) ;

			if( r[i].x1 >= r[i].x2 || r[i].y1 >= r[i].y2 )
			{
				area = 0 ;  goto end ;
			}
		}

		t = r[1] ;

		for( i = 2; i <= N; i++ )
		{
			/*if( r[i].x2 <= t.x1 || t.x2 <= r[i].x1 || r[i].y2 <= t.y1 || t.y2 <= r[i].y1 )
			{
				area = 0 ; break ;
			}*/

			t.x1 = r[i].x1 > t.x1 ? r[i].x1 : t.x1 ;
			t.x2 = r[i].x2 < t.x2 ? r[i].x2 : t.x2 ;

			t.y1 = r[i].y1 > t.y1 ? r[i].y1 : t.y1 ;
			t.y2 = r[i].y2 < t.y2 ? r[i].y2 : t.y2 ;

			if( t.x2 >= t.x1 && t.y2 >= t.x1 )
				area = ( t.x2 - t.x1 ) * ( t.y2 - t.y1 ) ;
			
			else {	area = 0 ; break ; }
		}

		end : ;
		printf( "Case %d: %d\n", kase++, area ) ;
	}
	return 0 ;
}

[rio]

Try this.
Input:

Code: Select all

2
2
1 1 0 0
0 0 1 1
1
0 0 1 1

-----
Rio