poj 1696:Space Ant（计算几何，凸包变种，极角排序）

Space Ant

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2876		Accepted: 1839

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

---

 

　　计算几何，凸包的变种，叉积的应用，极角排序。

　　题意是给你n个点，一只蚂蚁要从y坐标最小的那个点开始，逆时针依次走过所有的点，要求不能走之前走过的点，蚂蚁只能逆时针前进。

　　思路：由题意可知，蚂蚁就是走了一个螺旋状的路径，由于只能走逆时针，所以每次选择点的时候，就从剩下的点中选择最靠右的点（叉积比较），以此类推，到最后就能选择出所有的点。

　　代码：

 #include <iostream>
 using namespace std;
 #define eps 1e-10
 struct Point{    //定义点
     double x,y;
     Point(double x=,double y=):x(x),y(y) {}
 };
 typedef Point Vector;    //重定义向量
 Vector operator + (Vector a,Vector b)    //向量+向量
 {
     return Vector(a.x+b.x,a.y+b.y);
 }
 Vector operator - (Point a,Point b)    //点-点 = 向量
 {
     return Vector(a.x-b.x,a.y-b.y);
 }
 double Cross(Vector a,Vector b)    //求叉积
 {
     return a.x*b.y-b.x*a.y;
 }
 int main()
 {
     int T;
     cin>>T;
     while(T--){
         int n,i;
         Point p[];
         bool isw[] = {};    //记录走没走过
         cin>>n;
         for(i=;i<=n;i++){    //输入点集
             int t;
             cin>>t;
             cin>>p[t].x>>p[t].y;
         }
         //找出y坐标最小的那个点
         double yMin=p[].y;
         int num = ;
         for(i=;i<=n;i++){
             if(p[i].y < yMin){
                 yMin = p[i].y;
                 num = i;
             }
         }
         //构成凸包
         int pl[];    //存储结果
         int j;
         pl[] = num;
         isw[pl[]] = true;
         for(i=;i<n;i++){
             for(j=;j<=n;j++)    //选择一个没走过的点
                 if(!isw[j])
                     break;
             pl[i+] = j;
             for(j=;j<=n;j++){    //选择最右边的点
                 if(pl[i+]==j || isw[j])
                     continue;
                 if(Cross(p[j] - p[pl[i]],p[pl[i+]] - p[pl[i]])>)
                     pl[i+] = j;
             }
             isw[pl[i+]] = true;
         }
         cout<<n<<' ';
         for(i=;i<=n;i++)
             cout<<pl[i]<<' ';
         cout<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013