Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 4970 | Accepted: 3100 |
Description
- It can not turn right due to its special body structure.
- It leaves a red path while walking.
- 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
Output
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
It seems that this problem can be written directly in a polar order
Direct set of templates to write
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <algorithm> 5 #include <queue> 6 #include <map> 7 #include <vector> 8 #include <set> 9 #include <string> 10 #include <math.h> 11 12 using namespace std; 13 14 struct point { 15 double x, y; 16 int idx; 17 point() {} 18 point(int x, int y) : x(x), y(y) {} 19 point operator - (const point & a) const { 20 return point(x - a.x, y - a.y); 21 } 22 double operator ^ (const point & b) const { 23 return x * b.y - y * b.x; 24 } 25 double operator * (const point & a) const { 26 return x * a.x + y * a.y; 27 } 28 } p[110]; 29 double dist(point a, point b) { 30 return sqrt((a - b) * (a - b)); 31 } 32 int pos; 33 int cmp(point a, point b) { 34 double temp = (a - p[pos]) ^ (b - p[pos]); 35 if (temp == 0) return dist(p[pos], a) < dist(p[pos], b); 36 else if(temp < 0) return 0; 37 else return 1; 38 } 39 int main() { 40 int t, n; 41 scanf("%d", &t); 42 while(t-- ) { 43 scanf("%d", &n); 44 double x, y; 45 for (int i = 0 ; i < n ; i++) { 46 scanf("%d%lf%lf", &p[i].idx, &p[i].x, &p[i].y); 47 if (p[i].y < p[0].y || p[i].y == p[0].y && p[i].x < p[0].x) swap(p[i], p[0]); 48 } 49 pos = 0; 50 for (int i = 1 ; i < n ; i++) { 51 sort(p + i, p + n, cmp); 52 pos++; 53 } 54 printf("%d", n); 55 for (int i = 0 ; i < n ; i++) 56 printf(" %d", p[i].idx); 57 printf("\n"); 58 } 59 return 0; 60 }