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 D

The Art Gallery

Input: standard input

Output: standard output

 

Century Arts has hundreds of art galleries scattered all around the country and you are hired to write a program that determines whether any of the galleries has a critical point. The galleries are polygonal in shape and a critical point is a point inside that polygon from where the entire gallery is not visible.

 

For example, in gallery 1 (drawn below) point A is a critical point, but gallery 2 has no critical point at all.

                  

 

The Century Arts authorities will provide you with the shape of a gallery as a sequence of (x, y) co-ordinates (determined using a suitable origin) of the consecutive corner points of that gallery.

 

Input

The input file consists of several data blocks. Each data block describes one gallery.

 

The first line of a data block contains an integer N (3 £ N £ 50) indicating the number of corner points of the gallery. Each of the next N lines contains two integers giving the (x, y) co-ordinates of a corner point where 0 £ x, y £ 1000. Starting from the first point given in the input the corner points occur in the same order on the boundary of the gallery as they appear in the input. No three consecutive points are co-linear.

 

The input file terminates with a value of 0 for N.

 

Output

For each gallery in the input output the word "Yes" if the gallery contains a critical point, otherwise output the word "No". Each output must be on a separate line.

 

Sample Input

4
0 0
3 0
3 3
0 3
4
0 0
3 0
1 1
0 3
0

Sample Output

No
Yes

---

已經給定順序給點, 檢查外積是不是全部為正, 或者是全部為負


#include <stdio.h>
typedef struct {
    int x, y;
} Pt;
int 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 main() {
    int n, i;
    Pt D[100];
    while(scanf("%d", &n), n) {
        for(i = 0; i < n; i++)
            scanf("%d %d", &D[i].x, &D[i].y);
        D[n] = D[0], D[n+1] = D[1];
        if(cross(D[0], D[1], D[2]) >= 0) {
            for(i = 0; i < n; i++)
                if(cross(D[i], D[i+1], D[i+2]) < 0)
                    break;
        } else {
            for(i = 0; i < n; i++)
                if(cross(D[i], D[i+1], D[i+2]) > 0)
                    break;
        }
        if(i != n)
            puts("Yes");
        else
            puts("No");
    }
    return 0;
}