POJ 2007 Scrambled Polygon(凸包)

标签：poj 2007 scrambled p   凸包   

Description

A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the vertices of the polygon. When one starts at any vertex of a closed polygon and traverses each bounding line segment exactly once, one comes back to the starting vertex. 

A closed polygon is called convex if the line segment joining any two points of the polygon lies in the polygon. Figure 1 shows a closed polygon which is convex and one which is not convex. (Informally, a closed polygon is convex if its border doesn‘t have any "dents".) 

The subject of this problem is a closed convex polygon in the coordinate plane, one of whose vertices is the origin (x = 0, y = 0). Figure 2 shows an example. Such a polygon will have two properties significant for this problem. 

The first property is that the vertices of the polygon will be confined to three or fewer of the four quadrants of the coordinate plane. In the example shown in Figure 2, none of the vertices are in the second quadrant (where x < 0, y > 0). 

To describe the second property, suppose you "take a trip" around the polygon: start at (0, 0), visit all other vertices exactly once, and arrive at (0, 0). As you visit each vertex (other than (0, 0)), draw the diagonal that connects the current vertex with (0, 0), and calculate the slope of this diagonal. Then, within each quadrant, the slopes of these diagonals will form a decreasing or increasing sequence of numbers, i.e., they will be sorted. Figure 3 illustrates this point. 
 

Input

Output

Sample Input

0 0
70 -50
60 30
-30 -50
80 20
50 -60
90 -20
-30 -40
-10 -60
90 10

Sample Output

(0,0)
(-30,-40)
(-30,-50)
(-10,-60)
(50,-60)
(70,-50)
(90,-20)
(90,10)
(80,20)
(60,30)

Source

题意：给出若干个点，求出它们的凸包，并且按原点为第一点的逆时针方向输出。

#include<cstring>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#define PI 3.1415926

const double eps=1e-12;
#define _sign(x) ((x)>eps?1:((x)<-eps?2:0))

using namespace std;
const int MAXN = 1010;

struct Point {
    double x,y;
    Point() {}
    Point(double _x,double _y) {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const {
        return Point(x - b.x,y - b.y);
    }
//叉积
    double operator ^(const Point &b)const {
        return x*b.y - y*b.x;
    }
//点积
    double operator *(const Point &b)const {
        return x*b.x + y*b.y;
    }
//绕原点旋转角度B（弧度值），后x,y的变化
    void transXY(double B) {
        double tx = x,ty = y;
        x = tx*cos(B) - ty*sin(B);
        y = tx*sin(B) + ty*cos(B);
    }
};
Point L[MAXN];
int Stack[MAXN],top;
int n;
double m;

double dist(Point a,Point b) {
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int sgn(double x) {
    if(fabs(x)<eps)return 0;
    if(x<0)return -1;
    return 1;
}
//相对于L[0]的极角排序
bool _cmp(Point p1,Point p2) {
    double tmp = (p1-L[0])^(p2-L[0]);
    if(sgn(tmp) > 0)return true;
    else if(sgn(tmp) == 0 && sgn(dist(p1,L[0]) - dist(p2,L[0])) <= 0)
        return true;
    else return false;
}
void Graham(int n) {
    Point p0;
    int k = 0;
    p0 = L[0];
//找最下边的一个点
    for(int i = 1; i < n; i++) {
        if( (p0.y > L[i].y) || (p0.y == L[i].y && p0.x > L[i].x) ) {
            p0 = L[i];
            k = i;
        }
    }
    swap(L[k],L[0]);
    sort(L+1,L+n,_cmp);
    if(n == 1) {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2) {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return ;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2; i < n; i++) {
        while(top > 1 && sgn((L[Stack[top-1]]-L[Stack[top-2]])^(L[i]-L[Stack[top-2]])) <= 0)
            top--;
        Stack[top++] = i;
    }
}

int main() {
    //freopen("test.in","r",stdin);
    n=0;
    while(~scanf("%lf%lf",&L[n].x,&L[n].y)) {
        n++;
    }
    Graham(n);
    double ans=0;
    int p=0;
    for(int i=0; i<top; i++) {
        int u=Stack[i];
        if(L[u].x==0&&L[u].y==0) {
            p=i;
            break;
        }
    }
    for(int i=p; i<top; i++)
        printf("(%.0f,%.0f)\n",L[Stack[i]].x,L[Stack[i]].y);
    for(int i=0; i<p; i++) {
        printf("(%.0f,%.0f)\n",L[Stack[i]].x,L[Stack[i]].y);
    }
    return 0;
}

POJ 2007 Scrambled Polygon(凸包)

标签：poj 2007 scrambled p   凸包   

原文地址：http://blog.csdn.net/acm_baihuzi/article/details/47430595