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

The Great Divide

Input: standard input

Output: standard output

Time Limit: 8 seconds

Memory Limit: 32 MB

Somewhere in Gaul, there is a little village very like the village where Asterix and Obelix live. Not very long ago they had only one chief Altruistix and peace reigned in the village. But now those happy days are just dreams. The villagers are now divided. Some of the villagers have elected Majestix as their chief and the others have elected Cleverdix.

Majestix                                              Cleverdix

The two chiefs have decided to divide the village into two parts by digging a straight ditch through the middle of the village so that the houses of the supporters of Majestix lie on one part and those of the followers of Cleverdix lie on the other. So, they have invited Getafix, the venerable druid of Asterix’s village, to figure out whether such a dividing line exists or not.

Getafix

Since Getafix knows that you are so good in programming, he seeks your help.

Input

 The input may contain multiple test cases.

The first line of each test case contains two integers M and C (1 £ M, C £ 500), indicating the number of houses of the supporters of Majestix and Cleverdix respectively.

Each of the next M lines contains two integers x and y (-1000 £ x, y £ 1000) giving the co-ordinates of the house of a supporter of Majestix. For convenience each house is considered as a single point on the plane.

Each of the next C lines contains two integers x and y (-1000 £ x, y £ 1000) giving the co-ordinates of the house of a supporter of Cleverdix.

The input will terminate with two zeros for M and C.

Output

For each test case in the input output a line containing either “Yes” or “No” depending on whether there exists a straight line that divides the two set of houses. The dividing line can NOT contain points of both sides.

Sample Input

4 3
100 600
200 400
600 500
300 700
400 100
600 200
500 300

4 3
100 600
400 100
600 200
500 300
200 400
600 500
300 700

0 0

Sample Output

Yes
No

(World Final Warm-up Contest, Problem Setter: Rezaul Alam Chowdhury)

2. 檢查是否另一個凸包在另一個凸包內

3. 檢查凸包的邊是否有相交

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
using namespace std;
struct Pt {
    double x, y;
};
#define eps 1e-8
double cross(Pt o, Pt a, Pt b) {
    return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
int inPolygon(Pt p[], Pt q, int n) {
    if(n == 1)  return 0; // false
    if(n == 2) { // on segment
        if(p[0].x > q.x != p[1].x > q.x &&
           p[0].y > q.y != p[1].y > q.y &&
           fabs(cross(p[0], p[1], q)) < eps)
           return 1;
        return 0;
    }
    int i, j, k = 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)
           k++;
    }
    return k&1;
}
bool cmp(Pt a, Pt b) {
    if(a.x != b.x)
        return a.x < b.x;
    return a.y < b.y;
}
int monotone(Pt p[], int n, Pt ch[]) {
    sort(p, p+n, cmp);
    int i, m = 0, t;
    for(i = 0; i < n; i++) {
        while(m >= 2 && cross(ch[m-2], ch[m-1], p[i]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    for(i = n-1, t = m+1; i >= 0; i--) {
        while(m >= t && cross(ch[m-2], ch[m-1], p[i]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    return m-1;
}
int intersection(Pt as, Pt at, Pt bs, Pt bt) {
    if(cross(as, at, bs)*cross(as, at, bt) < 0 &&
       cross(at, as, bs)*cross(at, as, bt) < 0 &&
       cross(bs, bt, as)*cross(bs, bt, at) < 0 &&
       cross(bt, bs, as)*cross(bt, bs, at) < 0)
        return 1;
    return 0;
}
int M, C, i, j, k, l;
Pt Mp[505], Cp[505], Mch[1005], Cch[1005];
void sol() {
    int Mn = monotone(Mp, M, Mch);
    int Cn = monotone(Cp, C, Cch);
    for(i = 0; i < Cn; i++) {
        if(inPolygon(Mch, Cch[i], Mn)) {
            puts("No");
            return;
        }
    }
    for(i = 0; i < Mn; i++) {
        if(inPolygon(Cch, Mch[i], Cn)) {
            puts("No");
            return;
        }
    }
    for(i = 0, j = Mn-1; i < Mn; j = i++) {
        for(k = 0, l = Cn-1; k < Cn; l = k++) {
            if(intersection(Mch[i], Mch[j], Cch[k], Cch[l])) {
                puts("No");
                return;
            }
        }
    }
    puts("Yes");
}
int main() {
    while(scanf("%d %d", &M, &C) == 2 && M) {
        for(i = 0; i < M; i++)
            scanf("%lf %lf", &Mp[i].x, &Mp[i].y);
        for(i = 0; i < C; i++)
            scanf("%lf %lf", &Cp[i].x, &Cp[i].y);
        sol();
    }
    return 0;
}