poj 1696 Space Ant（极角排序）

Space Ant

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3661		Accepted: 2281

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

题意：一种生物只能逆时针走，且路线不与走过的路线交叉～

现在有好多植物，这种生物一次会走到一棵植物上，直到按照前面的行动规则没有植物可以走，结束。

那么问最长的依次拜访点的路线是怎么样～

恩，不废话了。

输入：t组数据，每组n个点，接下来每行三个数据分别代表 点的编号，点的横坐标，点的纵坐标（点为植物）。

输出：n个点，以及排过序的点的编号，

第一道极角排序～

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <math.h>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <set>
#define ll long long
using namespace std;

const double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point
{
    double x,y;
    int th;
    Point() {}
    Point(double _x,double _y)
    {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double  operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};

double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
int pos;
Point p[105];
bool cmp(Point a,Point b)
{
    double tmp = (a - p[pos])^(b - p[pos]);//叉乘判断方向
    if(sgn(tmp) == 0) return dist(a,p[pos]) < dist(b,p[pos]);//叉乘结果相同，则让距离远的优先在前面
    else if(sgn(tmp) < 0) return false;//刚好为逆时针，不交换
    else return true;//否则交换
}

int main(void)
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d %lf %lf",&p[i].th,&p[i].x,&p[i].y);
            if( p[i].y < p[0].y || (p[i].y == p[0].y && p[i].x < p[0].x) )
                swap(p[0],p[i]);//找到最左下端的点
        }
        pos = 0;
        for(int i = 0 ; i < n; i++)
        {
            sort(p+i,p+n,cmp);//每次都找出未排序中最外围的点
            pos++;
        }
        printf("%d",n);
        for(int i = 0; i < n; i++)
            printf(" %d",p[i].th);
        printf("
");
    }
    return 0;
}

ps:奉上几种极角排序的方法，虽然还没有用过～可以去看看 poj 2007

介绍几种极角排序：

1.利用叉积的正负来作cmp.(即是按逆时针排序).此题就是用这种方法

1 bool cmp(const point &a, const point &b)//逆时针排序 
2 {
3     point origin;
4     origin.x = origin.y = 0;
5     return cross(origin,b,origin,a) < 0;
6 }

2.利用complex的内建函数。

 1 #include<complex>
 2 #define x real()
 3 #define y imag()
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 bool cmp(const Point& p1, const Point& p2)
 8 {
 9     return arg(p1) < arg(p2);
10 }

3.利用arctan计算极角大小。（范围『-180，180』）

1 bool cmp(const Point& p1, const Point& p2)
2 {
3     return atan2(p1.y, p1.x) < atan2(p2.y, p2.x);
4 }

4.利用象限加上极角，叉积。

 1 bool cmp(const point &a, const point &b)//先按象限排序，再按极角排序，再按远近排序 
 2 {
 3     if (a.y == 0 && b.y == 0 && a.x*b.x <= 0)return a.x>b.x;
 4     if (a.y == 0 && a.x >= 0 && b.y != 0)return true;
 5     if (b.y == 0 && b.x >= 0 && a.y != 0)return false;
 6     if (b.y*a.y <= 0)return a.y>b.y;
 7     point one;
 8     one.y = one.x = 0;
 9     return cross(one,a,one,b) > 0 || (cross(one,a,one,b) == 0 && a.x < b.x);
10 }