poj 1696:Space Ant(计算几何,凸包变种,极角排序)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 2876 | Accepted: 1839 |
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
Source
#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
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