Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

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
*/