題目連結：https://codeforces.com/gym/102219/problem/H
部落格園食用連結：https://blog.csdn.net/qq_43906000/article/details/107884811

Recently, the nation was shocked by news of Sungai Kim Kim incident in Pasir Gudang, Johor, which has been polluted by chemical waste. Thousands of people who are affected had experienced nausea, dizziness and vomiting, and more than 100 schools were ordered to shut. In order to ensure that such incident will not happen again, an early warning system need o be developed so that residents can make early preparation, and authorities are able to move and act much faster.

A group of scientists has formed a committee to handle the incident, and a smart system with Internet of Things (IoT) sensors was suggested. Numerous sensors which can sense and monitor damages to the environment, either air or water, have been strategically installed around the state, and their coordinates are also recorded. However, the proposed system encountered a failure during its first testing phase. They want you to fix the problem in determining whether the given coordinates of sensors are safe or in the affected areas.

An affected area is defined as the polygon with the minimum length perimeter that can enclose all sensors that trigger warning signal within that area. For example, the sensors (represented by dots) of an affected area and its associated polygon, as well as safe (represented by triangles) and unsafe (represented by diamonds) points of the first dataset are illustrated below.

Input
The input will contain records of data for several test cases of affected areas. The first line of each data set contains a non-negative integer T, the number of test cases \((1≤T≤50)\). Each test case starts with two non-negative integer C and P which is the number of coordinates \((3≤C≤50)\), and points \((1≤P≤50)\), respectively. The next C lines contain coordinates (x-coordinate, y-coordinate) of each installed sensor, separated with blank spaces. The following P lines contain coordinates (x-coordinate, y-coordinate) of certain locations in the state, separated with blank spaces. All coordinates are integers between −500 and 500 inclusive.

Output
For each test case, output the following item:

First line: The number of the test cases. The first record corresponds to Case1, the second to Case2 , etc.

Next line: A listing of all the points that appear on the perimeter of the affected area. The points must be identified in the standard form "x-coordinate y- coordinate". The listing must be oriented counter-clockwise and begin and end with the same point.

Last line: For each point of location in the data set, output the line:

x−coordinatey−coordinateisstatus
where x−coordinatey−coordinate is the coordinate of the location from the input and status is ′safe′ or ′unsafe′. A location is considered unsafe it is within the sensor perimeter. A point in exactly at the edge of the perimeter is considered safe.

Each test case must be separated by an empty line. See example.

Example
Input
2
6 5
-477 -180
31 -266
-474 28
147 49
323 -53
277 -79
346 488
-139 -183
-427 129
386 -222
-408 -315
5 2
-52 -325
104 420
315 356
-192 8
493 146
404 228
-239 484

Output
Case 1
-477 -180
31 -266
323 -53
147 49
-474 28
-477 -180
346 488 is safe!
-139 -183 is unsafe!
-427 129 is safe!
386 -222 is safe!
-408 -315 is safe!

Case 2
-192 8
-52 -325
493 146
315 356
104 420
-192 8
404 228 is unsafe!
-239 484 is safe!

題目大意：給你n個點，你現在要從其中選擇一些點構成一個凸包，再給你m個點，判斷是否再凸包內，如果不在，則輸出\(x\ y\ is\ safe!\)否則輸出\(x\ y\ is\ unsafe!\)，注意當點再邊上的時候也認為他們是安全的，凸包的輸出要求逆時針方向且首位相接。

emmm，就是個裸的板子題。。。要是早點套一下zzq的板子就好了，網上的一堆板子全都WA1。。。。。好像用整形寫也是WA1，用double好像就過了。。。

套一下求凸包的板子，這個可以不要考慮順序的，反正最後你控制一些輸出順序就好了，至於判斷點是否再凸包內，這50的資料範圍量，隨便寫都OK，寫個\(O(n)\)的都可以，不過大多數的人手中可能都只有\(O(logn)\)的板子，也沒什麼可說的了，就是個板子模組

以下是AC程式碼：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long ll;
const int MAXN = 1e3+10;
const double esp=1e-8;

struct Point {
    double x,y;
    Point(double x_ = 0, double y_ = 0) : x(x_), y(y_) {}
    bool operator < (const Point& s) const {
        return x != s.x ? x < s.x : y < s.y;
    }
    Point operator - (const Point &A) const{
        return Point{x-A.x,y-A.y};
    }
} p[MAXN], ch[MAXN];

int dcmp(double x) { //返回x的正負
    if (fabs(x)<esp) return 0;
    return x<0?-1:1;
}

double Cross(Point A,Point B) { //叉積
    return A.x*B.y-A.y*B.x;//如果改成整形記得加LL
}

int Graham(Point *P,Point *R,int n) {
    int m = 0, k;
    sort(P, P + n);
    for(int i = 0; i < n; i++) {
        while(m > 1 && dcmp(Cross(R[m - 1]-R[m-2],P[i]-R[m - 2]))<= 0) m--;
        R[m++] = P[i];
    }
    k = m;
    for(int i = n - 2; i >= 0; i--) {
        while(m > k && dcmp(Cross(R[m - 1]-R[m-2],P[i]-R[m - 2]))<= 0) m--;
        R[m++] = P[i];
    }
    if(n > 1) m--;
    return m;
}

int n,m;
Point use[MAXN];

int IsPointInPolygon(Point P,Point *R){   //0外部，-1在邊上，1在內部
    int l=1, r=n-2,mid;
    while(l<=r){
        mid = (l+r) >> 1;
        int t1 = dcmp(Cross(R[mid]-R[0],P-R[0]));
        int t2 = dcmp(Cross(R[mid+1]-R[0],P-R[0]));
        if(t1>=0 &&t2<=0){
            int t3=dcmp(Cross(R[mid]-P,R[mid+1]-P));
            if(t3 < 0) return 0;
            else if((t1==0&&mid==1)||(t2==0&&mid+1==n-1)||t3==0)return -1;
            return 1;
        }
        else if(t1 < 0)r = mid -1;
        else l = mid+1;
    }
    return 0;
}

int main() {
    //freopen("in.txt","r",stdin);
    int T;
    scanf ("%d",&T);
    for (int cae=1; cae<=T; cae++) {
        scanf ("%d%d",&n,&m);
        for (int i=0; i<n; i++)
            scanf ("%lf%lf",&p[i].x, &p[i].y);
        for (int i=1; i<=m; i++)
            scanf ("%lf%lf",&use[i].x, &use[i].y);
        n = Graham(p, ch, n);
       // n--;
        printf("Case %d\n",cae);
        for (int i = 0; i < n; i++)
            printf("%.0f %.0f\n", ch[i].x, ch[i].y);
        printf("%.0f %.0f\n",ch[0].x,ch[0].y);

        for (int i=1; i<=m; i++){
            printf("%.0f %.0f is ",use[i].x,use[i].y);
            int ok=IsPointInPolygon(use[i],ch);
            if (ok==1) printf("unsafe!\n");
            else printf("safe!\n");
        }
        if (cae!=T) printf("\n");
    }
    return 0;
}