USACO 2021 February Contest, Bronze Problem 3. Clockwise Fence

USACO 2021 February Contest, Bronze Problem 3. Clockwise Fence

The fence surrounding Farmer John's largest pasture has fallen into disrepair, and he has finally decided to replace it with a new fence.

Unfortunately, as Farmer John is laying out the new fence, a large bee ends up chasing him around the pasture, and as a result, the fence ends up following a rather irregular path. The fence can be described by a string of characters, each either "N" (north), "E" (east), "S" (south), or "W" (west). Each character describes a 1-meter run of the fence. For example, if the string is NESW, this means the fence starts by moving north for 1 meter, then east for 1 meter, then south for 1 meter, then west for 1 meter, returning to its starting point.

The fence ends at the position where it started, and this is the only point visited more than once by the path of the fence (and the starting point is only re-visited once, at the end). As a result, the fence does indeed enclose a single connected region of the grassy pasture, even though this region could have a rather strange shape.

Farmer John is curious if the path in which he laid the fence traveled clockwise (with the enclosed region on the right side of the fence as one walks along the path of the fence in the order specified by the string) or counter-clockwise (with the enclosed region on the left side of the fence).

 

 

 

SAMPLE INPUT:

2
NESW
WSSSEENWNEESSENNNNWWWS

SAMPLE OUTPUT:

CW
CCW

The two fence paths with @ denoting the starting point:

 

*>*
^ v
@<*

  *<*<*<*
  v     ^
*<@     *
v       ^
* *>*>* *
v ^   v ^
* *<* * *
v   ^ v ^
*>*>* *>*

 

Problem credits: Brian Dean

USACO 2021 February Contest, Bronze Problem 3. Clockwise Fence 题解(翰林国际教育提供，仅供参考)

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(Analysis by Dhruv Rohatgi and Brian Dean )

Intuitively, a clockwise fence will "tend" to turn right, and a counterclockwise fence will tend to turn left. More concretely, for every two adjacent fence segments, one can compute the angle turned at that corner: either no turn (if the two segments are in the same direction), or a turn by 90∘90∘ counterclockwise (e.g. if the first segment is E and the second is N), or a turn by −90∘−90∘ counterclockwise (e.g. if the first segment is E and the second is S). Doing a few examples by hand will reveal that the sum of these angles, over all corners of the fence, is precisely 360∘360∘ if the fence is counterclockwise, and −360∘−360∘ if clockwise (other sums can be achieved if the fence is allowed to intersect itself, but we don't have to worry about this complication in this problem).

This fact can be proven a variety of ways, such as by inducting on the area of the region enclosed by the fence. It gives a linear-time algorithm, shown below.

Dhruv's code:

 

#include <iostream>
#include <string>
#include <cassert>
using namespace std;

int angle_from_direction(char a)
{
	if(a == 'E') return 0;
	if(a == 'N') return 90;
	if(a == 'W') return 180;
	if(a == 'S') return 270;
}

int angle_change(char a,char b)
{
	int theta1 = angle_from_direction(a);
	int theta2 = angle_from_direction(b);
	if(theta2 == (theta1 + 90)%360) return 90;
	else if(theta2 == theta1) return 0;
	else if(theta2 == (theta1 + 270)%360) return -90;
	else assert(false);	//fence should not backtrack on itself
}

void test(string s)
{
	int total_change = 0;
	for(int i=0;i<s.size();i++)
		total_change += angle_change(s[i],s[(i+1)%s.size()]);
	if(total_change == 360) cout << "CCW\n";
	else cout << "CW\n";
}

int main()
{
	int N;
	string s;
	cin >> N;
	for(int i=0;i<N;i++)
	{
		cin >> s;
		test(s);
	}
}

There are several other ways to effectively approach this problem. For example, one can look at the direction of any "boundary" segment. That is, look any horizontally-oriented segment that is as far north as possible --- if it is directed east, the entire path must be clockwise, and if directed west, the entire path must be counterclockwise. In fact, we can draw any horizontal or vertical line through the scene and deduce from the direction of the segments along this cross section the overall orientation of the fence. For example, if we draw a horizontal line through the scene, it will be cut by some vertical segments pointing north and some pointing south (in fact, these will alternate in direction along the cross section); if the westmost segment points north, the entire path has a clockwise orientation, and vice versa.

The shoelace formula for computing areas of polygons also leads to a solution for this problem, since it delivers a "signed" area, being positive or negative depending on the orientation of the polygon in question.
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