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 F
The Sultan’s Problem
Input: Standard Input

Output: Standard Output

Once upon a time, there lived a greatsultan, who was very much fond of his wife. He wanted to build a Tajmahal for his wife (ya, oursultan idolized Mughal emperor Shahjahan).But alas, due to budget cuts, loans, dues and many manythings, he had no fund to build something so big. So, he decided to build abeautiful garden, inside his palace.

The garden is a rectangular pieceof land. All the palace’s water lines run through the garden, thus, dividing itinto pieces of varying shapes. He wanted to cover each piece of the land withflowers of same kind.

Figure 1

The figure above shows a sampleflower garden. All the lines are straight lines, with two end points on twodifferent edge of the rectangle. Each piece of the garden is covered with thesame kind of flowers.

The garden has a small fountain,located at position (x,y).You can assume that, it is not on any of the lines. He wants to fill that piecewith her favourite flower. So, he asks you to findthe area of that piece.

Input

Input contains around 500 testcases. Each case starts with 5 integers, N, W, H, x,y, the number of lines, width and height of the garden and the locationof the fountain. Next N lines each contain 4 integers, x₁y₁ x₂ y₂, the two end points of the line.The end points are always on the boundary of the garden. You can assume that,the fountain is not located on any line.

Output

For each test case, output thecase number, with the area of the piece covered with her favouriteflower. Print 3 digits after the decimal point.

Constraints

• 1 ≤ W, H ≤ 200,000
• 1 ≤ N ≤ 500

Problemsetter: Manzurur Rahman Khan

Special Thanks To: Mohammad MahmudurRahman

解題思路：

一個型態 polygon，以及求一條長直線交多邊形的兩交點，
如果有交的話，將這個多邊形一分為二，同時只保留涵蓋花園的那塊多邊形，
最後使用行列式算花園面積。

#include <stdio.h>
#include <math.h>
struct Pt {
    double x, y;
};
struct Polygon {
    Pt p[505];
    int n;
};
double calcArea(Polygon &p) {
    static int i;
    double sum = 0;
    p.p[p.n] = p.p[0];
    for(i = 0; i < p.n; i++)
        sum += p.p[i].x*p.p[i+1].y - p.p[i].y*p.p[i+1].x;
    return fabs(sum/2);
}
void print(Polygon &p) {
    for(int i = 0; i < p.n; i++)
        printf("%lf %lf\n", p.p[i].x, p.p[i].y);
    puts("=====");
}
int main() {
    int n, w, h;
    int cases = 0;
    int i, j, k;
    double sx, sy, ex, ey, xi, yi;
    while(scanf("%d %d %d", &n, &w, &h) == 3) {
        scanf("%lf %lf", &xi, &yi);
        Polygon A, B, C;
        A.n = 4;
        A.p[0].x = 0, A.p[0].y = 0;
        A.p[1].x = w, A.p[1].y = 0;
        A.p[2].x = w, A.p[2].y = h;
        A.p[3].x = 0, A.p[3].y = h;
        while(n--) {
            scanf("%lf %lf %lf %lf", &sx, &sy, &ex, &ey);
            // ax + by + c = 0
            double a, b, c;
#define eps 1e-8
            double m = (sy-ey)/(sx-ex);
            if(fabs(sx-ex) < eps)
                a = 1, b = 0, c = -sx;
            else
                a = m, b = -1, c = -(sx*a+b*sy);
            //printf("%lf x + %lf y + %lf = 0\n", a, b, c);
            A.p[A.n] = A.p[0];
            B.n = 0, C.n = 0;
            Pt intP[2];
            int point = 0;
            for(i = 0; i < A.n; i++) {
                if(point == 1) {
                    if(B.n == 0 || fabs(B.p[B.n-1].x-A.p[i].x) > eps ||
                       fabs(B.p[B.n-1].y-A.p[i].y) > eps)
                    B.p[B.n++] = A.p[i];
                } else {
                    if(C.n == 0 || fabs(C.p[C.n-1].x-A.p[i].x) > eps ||
                       fabs(C.p[C.n-1].y-A.p[i].y) > eps)
                    C.p[C.n++] = A.p[i];
                }
                //printf("(%lf %lf)-(%lf %lf)\n", A.p[i].x, A.p[i].y, A.p[i+1].x, A.p[i+1].y);
                if((a*A.p[i].x+b*A.p[i].y+c)*(a*A.p[i+1].x+b*A.p[i+1].y+c) <= eps) {
                    if(point == 2)  continue;
                    double ta, tb, tc;
                    double tm = (A.p[i].y-A.p[i+1].y)/(A.p[i].x-A.p[i+1].x);
                    if(fabs(A.p[i].x-A.p[i+1].x) < eps)
                        ta = 1, tb = 0, tc = -A.p[i].x;
                    else
                        ta = tm, tb = -1, tc = -(A.p[i].x*ta+A.p[i].y*tb);
                    // ax+by+c = 0, ta*x+tb*y+tc = 0
                    //printf("%lf x + %lf y + %lf = 0\n", ta, tb, tc);
                    double rx, ry, r;
                    r = a*tb-ta*b;
                    rx = (-c)*tb-(-tc)*b;
                    ry = a*(-tc)-ta*(-c);
                    rx = rx/r;
                    ry = ry/r;
                    if(fabs(r) < eps)   continue; // no intersection
                    if(point == 1) {
                        if(fabs(rx-intP[0].x) < eps && fabs(ry-intP[0].y) < eps)
                            continue;
                    }
                    //printf("intersection %lf %lf\n", rx, ry);
                    intP[point].x = rx, intP[point].y = ry;
                    if(B.n == 0 || fabs(B.p[B.n-1].x-rx) > eps || fabs(B.p[B.n-1].y-ry) > eps)
                        B.p[B.n++] = intP[point];
                    if(C.n == 0 || fabs(C.p[C.n-1].x-rx) > eps || fabs(C.p[C.n-1].y-ry) > eps)
                        C.p[C.n++] = intP[point];
                    point++;
                }
            }
            if(point != 2)
                continue;
            int f1, f2;
            f1 = (a*B.p[B.n/2].x + b*B.p[B.n/2].y + c) < eps;
            f2 = (a*xi + b*yi + c) < eps;
            //print(B);
            //print(C);
            if(f1 == f2 && point) {// same side
                //puts("B side");
                A = B;
            } else {
                //puts("C side");
                A = C;
            }
            while(fabs(A.p[A.n-1].x-A.p[0].x) < eps &&
               fabs(A.p[A.n-1].y-A.p[0].y) < eps)
                A.n--;
            //print(A);
        }
        printf("Case #%d: %.3lf\n", ++cases, calcArea(A));
    }
    return 0;
}