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 C: Pick-up sticks

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.

Input consists of a number of cases. The data for each case start with 1 ≤ n ≤ 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.

The picture to the right below illustrates the first case from input.

Sample input

5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0

Output for sample input

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

---

Piotr Rudnicki


#include <stdio.h>
#define min(x, y) ((x) < (y) ? (x) : (y))
#define max(x, y) ((x) > (y) ? (x) : (y))
struct Point {
    double x, y;
};
struct Segment {
    Point s, t;
};
int in(double a, double b, double c) {
    return c >= a && c <= b;
}
int onLine(Segment& a, Point& c) {
    static double minx, maxx, miny, maxy;
    minx = min(a.s.x, a.t.x);
    maxx = max(a.s.x, a.t.x);
    miny = min(a.s.y, a.t.y);
    maxy = max(a.s.y, a.t.y);
    if(in(minx, maxx, c.x) != 0 && in(miny, maxy, c.y) != 0) {
        if((a.s.x-a.t.x)*(c.y-a.s.y) == (a.s.y-a.t.y)*(c.x-a.s.x)) {
            return 1;
        }
    }
    return 0;
}
double cross(Point &o, Point &a, Point &b) {
    return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
int intersection(Segment& a, Segment& b) {
    if(onLine(a, b.s) || onLine(a, b.t) || onLine(b, a.s) || onLine(b, a.t))
        return 1;
    if(cross(a.s, a.t, b.s)*cross(a.s, a.t, b.t) < 0 &&
       cross(a.t, a.s, b.s)*cross(a.t, a.s, b.t) < 0 &&
       cross(b.s, b.t, a.s)*cross(b.s, b.t, a.t) < 0 &&
       cross(b.t, b.s, a.s)*cross(b.t, b.s, a.t) < 0
       )
        return 1;
    return 0;
}
Segment stk[1005];
int stkin[1005];
int main() {
    int n, i, j, k;
    while(scanf("%d", &n) == 1 && n) {
        int idx = 0;
        Segment T;
        for(i = 0; i < n; i++) {
            scanf("%lf %lf %lf %lf", &T.s.x, &T.s.y, &T.t.x, &T.t.y);
            for(j = 0, k = 0; j < idx; j++)
                if(!intersection(T, stk[j]))
                    stkin[k] = stkin[j], stk[k++] = stk[j];
            idx = k;
            stkin[idx] = i, stk[idx++] = T;
        }
        printf("Top sticks:");
        for(i = 0; i < idx; i++) {
            if(i)   putchar(',');
            printf(" %d", stkin[i]+1);
        }
        puts(".");
    }
    return 0;
}