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 E: Hotter Colder

The children's game Hotter Colder is played as follows. Player A leaves the room while player B hides an object somewhere in the room. Player A re-enters at position (0,0) and then visits various other positions about the room. When player A visits a new position, player B announces "Hotter" if this position is closer to the object than the previous position; player B announces "Colder" if it is farther and "Same" if it is the same distance.

Input consists of up to 50 lines, each containing an x,y coordinate pair followed by "Hotter", "Colder", or "Same". Each pair represents a position within the room, which may be assumed to be a square with opposite corners at (0,0) and (10,10). For each line of input print a line giving the total area of the region in which the object may have been placed, to 2 decimal places. If there is no such region, output 0.00.

Sample Input

10.0 10.0 Colder
10.0 0.0 Hotter
0.0 0.0 Colder
10.0 10.0 Hotter

Output for Sample Input

50.00
37.50
12.50
0.00

---

題目說明：
平面上會有一個不知道的在哪的點(x, y)
玩家A從(0,0)出發，然後玩家A會走到下一個點 (nx, ny)，
玩家B會跟A講他這個點與上一個走過的點 (lx, ly)，
哪個離(x,y) 比較近，如果是 Hotter 代表 (nx, ny) 比較近，反之 (lx, ly)
如果是 Same 的話，直接輸出 0.000 直至最後。

題目分析：
這是一個半平面交的問題，將 (nx, ny) 與 (lx, ly) 做中垂線，
然後用中垂線將這個多邊形一劈兩半，保留其中的一塊，
如果是 Hotter 保留 (nx, ny) 同側的那塊，反之 (lx, ly)。

#include <stdio.h>
#include <math.h>
#include <string.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 = 10, h = 10;
    int i, j, k;
    double sx, sy, ex, ey, xi, yi;
    char cmd[50];
    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;
    sx = 0, sy = 0;
    while(scanf("%lf %lf", &ex, &ey) == 2) {
        scanf("%s", cmd);
        // ax + by + c = 0
        double a, b, c;
#define eps 1e-8
        double m = -(sy-ey)/(sx-ex);
        a = sx-ex, b = sy-ey;
        c = -(a*((sx+ex)/2)+b*((sy+ey)/2));
        //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) {
            printf("%.2lf\n", calcArea(A));
            sx = ex, sy = ey;
            continue;
        }
        int f1, f2;
        f1 = (a*B.p[B.n/2].x + b*B.p[B.n/2].y + c) < eps;
        if(!strcmp("Same", cmd))
            A.n = 0, B.n = 0, C.n = 0;
        else if(!strcmp("Hotter", cmd))
            xi = ex, yi = ey;
        else
            xi = sx, yi = sy;
        f2 = (a*xi + b*yi + c) < eps;
        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("%.2lf\n", calcArea(A));
        sx = ex, sy = ey;
    }
    return 0;
}