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

Allied Chute Manufacturers is a company that builds trash chutes. A trash chute is a hollow tube installed in buildings so that trash dropped in at the top will fall down and be collected in the basement. Designing trash chutes is actually highly nontrivial. Depending on what kind of trash people are expected to drop into them, the trash chute needs to have an appropriate size. And since the cost of manufacturing a trash chute is proportional to its size, the company always would like to build a chute that is as small as possible. Choosing the right size can be tough though.

We will consider a 2-dimensional simplification of the chute design problem. A trash chute points straight down and has a constant width. Objects that will be dropped into the trash chute are modeled as polygons. Before an object is dropped into the chute it can be rotated so as to provide an optimal fit. Once dropped, it will travel on a straight path downwards and will not rotate in flight. The following figure shows how an object is first rotated so it fits into the trash chute.

Your task is to compute the smallest chute width that will allow a given polygon to pass through.

Input 

The input contains several test cases. Each test case starts with a line containing an integer n (3n100), the number of points in the polygon that models the trash item.

The next n lines then contain pairs of integers x_i and y_i (0x_i, y_i10⁴), giving the coordinates of the polygon vertices in order. All points in one test case are guaranteed to be mutually distinct and the polygon sides will never intersect. (Technically, there is one inevitable exception of two neighboring sides sharing their common vertex. Of course, this is not considered an intersection.)

The last test case is followed by a line containing a single zero.

Output 

For each test case, display its case number followed by the width of the smallest trash chute through which it can be dropped. Display the minimum width with exactly two digits to the right of the decimal point, rounding up to the nearest multiple of 1/100. Answers within 1/100 of the correct rounded answer will be accepted.

Follow the format of the sample output.

Sample Input 

3
0 0
3 0
0 4
4
0 10
10 0
20 10
10 20
0

Sample Output 

Case 1: 2.40
Case 2: 14.15

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題目描述：
給定一個垃圾的多邊形，問丟入垃圾桶的最小寬度為何。

題目解法：

可以使用凸包，會比較快一點。
這題需要點線距的公式 ax+by+c = 0, 對於點 (x, y), 距離是 |ax+by+c|/sqrt(a*a+b*b)

如果不使用凸包，窮舉任兩點拉一直線，計算兩側的最大寬度加起來。
然後取最小值。O(n^3)

用凸包則是 O(nlogn + n*n)

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define eps 1e-8
struct Pt {
    double x, y;
    bool operator<(const Pt &A) const {
        if(fabs(x-A.x) > eps)
            return x < A.x;
        return y < A.y;
    }
};
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 monotone(Pt p[], int n, Pt ch[]) {
    sort(p, p+n);
    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 main() {
    /*freopen("in.txt", "r+t", stdin);
	freopen("out2.txt", "w+t", stdout);*/
    int n, m, i, j, k;
    int cases = 0;
    Pt D[505], ch[505];
    while(scanf("%d", &n) == 1 && n) {
        for(i = 0; i < n; i++)
            scanf("%lf %lf", &D[i].x, &D[i].y);
        m = monotone(D, n, ch);
        double ret = 1e+60;
        for(i = 0, j = m-1; i < m; j = i, i++) {
            double na = -(ch[i].y-ch[j].y), nb = ch[i].x-ch[j].x;
            double c = -(na*ch[i].x+nb*ch[i].y);
            double len = hypot(na, nb);
            double farth = 0;
            for(k = 0; k < n; k++)
                farth = max(farth, fabs(na*D[k].x+nb*D[k].y+c)/(len));
            ret = min(ret, farth);
        }
        printf("Case %d: %.2lf\n", ++cases, ret);
    }
    return 0;
}