[POJ 1696] Space Ant【计算几何】

  • 2017-12-11
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Problem:

Time Limit: 1000MS Memory Limit: 10000K

Description

The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y1999 and called it M11. It has only one eye on the left side of its head and just three feet all on the right side of its body and suffers from three walking limitations:

  1. It can not turn right due to its special body structure.
  2. It leaves a red path while walking.
  3. It hates to pass over a previously red colored path, and never does that.

The pictures transmitted by the Discovery space ship depicts that plants in the Y1999 grow in special points on the planet. Analysis of several thousands of the pictures have resulted in discovering a magic coordinate system governing the grow points of the plants. In this coordinate system with x and y axes, no two plants share the same x or y.
An M11 needs to eat exactly one plant in each day to stay alive. When it eats one plant, it remains there for the rest of the day with no move. Next day, it looks for another plant to go there and eat it. If it can not reach any other plant it dies by the end of the day. Notice that it can reach a plant in any distance.
The problem is to find a path for an M11 to let it live longest.
Input is a set of (x, y) coordinates of plants. Suppose A with the coordinates (xA, yA) is the plant with the least y-coordinate. M11 starts from point (0,yA) heading towards plant A. Notice that the solution path should not cross itself and all of the turns should be counter-clockwise. Also note that the solution may visit more than two plants located on a same straight line.

Input

The first line of the input is M, the number of test cases to be solved (1 <= M <= 10). For each test case, the first line is N, the number of plants in that test case (1 <= N <= 50), followed by N lines for each plant data. Each plant data consists of three integers: the first number is the unique plant index (1..N), followed by two positive integers x and y representing the coordinates of the plant. Plants are sorted by the increasing order on their indices in the input file. Suppose that the values of coordinates are at most 100.

Output

Output should have one separate line for the solution of each test case. A solution is the number of plants on the solution path, followed by the indices of visiting plants in the path in the order of their visits.

Sample Input

2
10
1 4 5
2 9 8
3 5 9
4 1 7
5 3 2
6 6 3
7 10 10
8 8 1
9 2 4
10 7 6
14
1 6 11
2 11 9
3 8 7
4 12 8
5 9 20
6 3 2
7 1 6
8 2 13
9 15 1
10 14 17
11 13 19
12 5 18
13 7 3
14 10 16

Sample Output

10 8 7 3 4 9 5 6 2 1 10
14 9 10 11 5 12 8 7 6 13 4 14 1 3 2

Source

Tehran 1999

Solution:

分析可知,使用凸包的卷包裹法(Gift Wrapping Method),每次取左转角度最小的点,总可以取到所有点

故本题的难点在于实现卷包裹法,利用叉积的传递性,可采用叉积判断左转角度最小的点。

Code: O(N2M) [192K, 0MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define eps 1e-8
#define dmax 1e100
#define sqr(x) ((x) * (x))
using namespace std;

int M, N;

#define Vector Point
struct Point{
	double x, y;
	
	Point() {}
	Point(double x, double y): x(x), y(y) {}
	
	inline Vector operator - (const Point p2) const {return Vector(x - p2.x, y - p2.y);}
	
	inline double to_Point(const Point p2) const {return sqrt(sqr(x - p2.x) + sqr(y - p2.y));}
};

inline double Cross(const Vector vec1, const Vector vec2) {return vec1.x * vec2.y - vec1.y * vec2.x;}

struct Y1999{
	int id;
	Point u;
} pl[55];

bool vis[55];

int main(){
	scanf("%d", &M);
	while(M--){
		scanf("%d", &N);
		Point cur(0, dmax);
		for(register int i = 1; i <= N; i++){
			scanf("%d%lf%lf", &pl[i].id, &pl[i].u.x, &pl[i].u.y);
			if(pl[i].u.y < cur.y) cur.y = pl[i].u.y;  // Find the min Y coordinate
		}
		memset(vis, 0, sizeof(vis));
		printf("%d", N);  // We can prove that all the plants can be eaten
		for(register int i = 1; i <= N; i++){
			int outer = 1;
			while(vis[outer]) outer++;
			for(register int i = outer + 1; i <= N; i++){
				if(vis[i]) continue;
				double crs = Cross(pl[i].u - cur, pl[outer].u - cur);
				if(crs > eps || (fabs(crs) <= eps && cur.to_Point(pl[i].u) < cur.to_Point(pl[outer].u))) outer = i;
				// Use gift wrapping method of convex hull
			}
			printf(" %d", outer);
			vis[outer] = 1, cur = pl[outer].u;
		}
		puts("");
	}
	return 0;
}

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OPEN AT 2017.12.10

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