Problem K

Polygon intersection

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

Given any two arbitrary polygons, determine their intersection if it exists.

Input

The input file contains several sets of input.

Each set will consist of two polygon description. Each polygon description begins with a positive integer n corresponding to the number of vertices, followed by n lines with each line containing a pair of integer (x,y) representing the x and y coordinates of a polygon vertex. The vertices will be given in clockwise traversal order and no two polygon edges would overlap. You should terminate your program if n < 3. You may assume all integer input is less than or equal to 100.

Output

For each set of input follow the follow the output description below.

If the two convex polygons do not intersect then print out 0. Otherwise, print out the number of intersecting points of the input polygons, followed by the points. The output must begin with the bottom leftmost point and must be listed in lexicographical ordering. Your answers must be rounded up to two digits after the decimal point

Sample Input:

3 

0 0 

0 2 

2 2

4 

5 5 

5 10 

10 10 

10 5

4 

1 1 

5 1 

5 5 

1 5

4 

3 0 

6 3 

3 6 

0 3

0

Sample Output

0

8 

1.00 2.00

1.00 4.00 

2.00 1.00 

2.00 5.00 

4.00 1.00 

4.00 5.00 

5.00 2.00 

5.00 4.00

(The Decider Contest, Source: Queens Univ & Univ of Toronto Local Contest)

找兩個多邊形邊的交點，以及多邊形的頂點是否在另一個多邊形內。
記得去重複的交點後輸出，如果有多算的可能。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define eps 1e-8
struct Pt {
    double x, y;
    bool operator<(const Pt &a) const {
        if(fabs(x-a.x) > eps)
            return x < a.x;
        return y < a.y;
    }
};
struct Seg {
    Pt s, e;
};
int inInterval(Seg a, Pt p) {
    return
        min(a.s.x, a.e.x) <= p.x &&
        p.x <= max(a.s.x, a.e.x) &&
        min(a.s.y, a.e.y) <= p.y &&
        p.y <= max(a.s.y, a.e.y);
}
int calcIntersection(Seg a, Seg b, Pt &p) {
    double a1, b1, c1, a2, b2, c2;
    double D, Dx, Dy;
    a1 = a.s.y-a.e.y, b1 = -a.s.x+a.e.x;
    a2 = b.s.y-b.e.y, b2 = -b.s.x+b.e.x;
    c1 = a1*a.s.x + b1*a.s.y;
    c2 = a2*b.s.x + b2*b.s.y;
    D = a1*b2 - a2*b1;
    Dx = c1*b2 - c2*b1;
    Dy = a1*c2 - a2*c1;
    if(fabs(D) < eps) // NONE or LINE
        return 0;
    p.x = Dx/D, p.y = Dy/D;
    /*printf("%lf %lf - %lf %lf\n", a.s.x, a.s.y, a.e.x, a.e.y);
    printf("%lf %lf - %lf %lf\n", b.s.x, b.s.y, b.e.x, b.e.y);
    printf("%lf %lf\n", p.x, p.y);*/
    return inInterval(a, p) == 1 && inInterval(b, p) == 1;
}
int inPolygon(Pt p[], int n, Pt q) {
    int i, j, cnt = 0;
    for(i = 0, j = n-1; i < n; j = i++) {
        if(p[i].y > q.y != p[j].y > q.y &&
           q.x < (p[j].x-p[i].x)*(q.y-p[i].y)/(p[j].y-p[i].y) + p[i].x)
           cnt++;
    }
    return cnt&1;
}
int main() {
    int n, m;
    int i, j, p, q;
    Pt a[105], b[105];
    while(scanf("%d", &n) == 1 && n) {
        for(i = 0; i < n; i++)
            scanf("%lf %lf", &a[i].x, &a[i].y);
        scanf("%d", &m);
        for(i = 0; i < m; i++)
            scanf("%lf %lf", &b[i].x, &b[i].y);
        Seg s1, s2;
        Pt ret[10005];
        int retN = 0;
        for(i = 0, j = n-1; i < n; j = i++) {
            s1.s = a[i], s1.e = a[j];
            for(p = 0, q = m-1; p < m; q = p++) {
                s2.s = b[p], s2.e = b[q];
                if(calcIntersection(s1, s2, ret[retN])) {
                    retN++;
                    /*printf("%lf %lf - %lf %lf\n", s1.s.x, s1.s.y, s1.e.x, s1.e.y);
                    printf("%lf %lf - %lf %lf +++\n", s2.s.x, s2.s.y, s2.e.x, s2.e.y);*/
                }
            }
        }
        for(i = 0; i < n; i++)
            if(inPolygon(b, m, a[i]))
                ret[retN++] = a[i];
        for(i = 0; i < m; i++)
            if(inPolygon(a, n, b[i]))
                ret[retN++] = b[i];
        if(retN == 0) {
            puts("0");
            continue;
        }
        sort(ret, ret+retN);
        for(i = 1, j = 0; i < retN; i++) {
            if(fabs(ret[i].x-ret[j].x) > eps ||
                fabs(ret[i].y-ret[j].y) > eps)
                ret[++j] = ret[i];
        }
        retN = j+1;
        printf("%d\n", retN);
        for(i = 0; i < retN; i++)
            printf("%.2lf %.2lf\n", ret[i].x, ret[i].y);
    }
    return 0;
}
/*
4
3 0
5 2
3 3
1 2
4
3 1
5 3
3 5
1 3
*/