POJ 1228 Grandpa's Estate（凸包唯一性判断）

Description

Being the only living descendant of his grandfather, Kamran the Believer inherited all of the grandpa's belongings. The most valuable one was a piece of convex polygon shaped farm in the grandpa's birth village. The farm was originally separated from the neighboring farms by a thick rope hooked to some spikes (big nails) placed on the boundary of the polygon. But, when Kamran went to visit his farm, he noticed that the rope and some spikes are missing. Your task is to write a program to help Kamran decide whether the boundary of his farm can be exactly determined only by the remaining spikes.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains an integer n (1 <= n <= 1000) which is the number of remaining spikes. Next, there are n lines, one line per spike, each containing a pair of integers which are x and y coordinates of the spike.

Output

There should be one output line per test case containing YES or NO depending on whether the boundary of the farm can be uniquely determined from the input.

 

题目大意：原本有一个凸多边形，现在我们不知道它的边，它的一部分点也有可能消失了，问用剩下的点能不能确定那个多边形。

思路：先求个凸包，然后判断是否凸包相邻的两点之间都存在一个题目给出的点使得这个点在这条边上。

因为有一个点在这个边上呢，这条边就确定下来了。

嘛其实没有这么麻烦，因为我在做模板所以……

啊，对了，全部点是一条直线的时候输出NO，我上了求多边形面积的模板←_←

 

代码（0MS）：

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

const int MAXN = 1010;
const double EPS = 1e-8;
const double PI = acos(-1.0);//3.14159265358979323846

inline int sgn(double x) {
    return (x > EPS) - (x < -EPS);
}

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y): x(x), y(y) {}
    void read() {
        scanf("%lf%lf", &x, &y);
    }
    bool operator == (const Point &rhs) const {
        return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0;
    }
    bool operator < (const Point &rhs) const {
        if(y != rhs.y) return y < rhs.y;
        return x < rhs.x;
    }
    Point operator + (const Point &rhs) const {
        return Point(x + rhs.x, y + rhs.y);
    }
    Point operator - (const Point &rhs) const {
        return Point(x - rhs.x, y - rhs.y);
    }
    Point operator * (const int &b) const {
        return Point(x * b, y * b);
    }
    Point operator / (const int &b) const {
        return Point(x / b, y / b);
    }
    double length() const {
        return sqrt(x * x + y * y);
    }
    Point unit() const {
        return *this / length();
    }
};
typedef Point Vector;

double dist(const Point &a, const Point &b) {
    return (a - b).length();
}

double cross(const Point &a, const Point &b) {
    return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn left
double cross(const Point &sp, const Point &ed, const Point &op) {
    return sgn(cross(sp - op, ed - op));
}

double area(const Point& a, const Point &b, const Point &c) {
    return fabs(cross(a - c, b - c)) / 2;
}

struct Seg {
    Point st, ed;
    Seg() {}
    Seg(Point st, Point ed): st(st), ed(ed) {}
    void read() {
        st.read(); ed.read();
    }
};
typedef Seg Line;

bool isOnSeg(const Seg &s, const Point &p) {
    return (p == s.st || p == s.ed) ||
        (((p.x - s.st.x) * (p.x - s.ed.x) < 0 ||
          (p.y - s.st.y) * (p.y - s.ed.y) < 0) &&
         sgn(cross(s.ed, p, s.st) == 0));
}

bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) {
    return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
        (max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
        (max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
        (max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
        (cross(s2, e1, s1) * cross(e1, e2, s1) >= 0) &&
        (cross(s1, e2, s2) * cross(e2, e1, s2) >= 0);
}

bool isIntersected(const Seg &a, const Seg &b) {
    return isIntersected(a.st, a.ed, b.st, b.ed);
}

bool isParallel(const Seg &a, const Seg &b) {
    return sgn(cross(a.ed - a.st, b.ed - b.st)) == 0;
}

//return Ax + By + C =0 's A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
    A = L.ed.y - L.st.y;
    B = L.st.x - L.ed.x;
    C = L.ed.x * L.st.y - L.st.x * L.ed.y;
}

Point intersection(const Line &a, const Line &b) {
    double A1, B1, C1;
    double A2, B2, C2;
    Coefficient(a, A1, B1, C1);
    Coefficient(b, A2, B2, C2);
    Point I;
    I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
    I.y =   (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
    return I;
}

bool isEqual(const Line &a, const Line &b) {
    double A1, B1, C1;
    double A2, B2, C2;
    Coefficient(a, A1, B1, C1);
    Coefficient(b, A2, B2, C2);
    return sgn(A1 * B2 - A2 * B1) == 0 && sgn(A1 * C2 - A2 * C1) == 0 && sgn(B1 * C2 - B2 * C1) == 0;
}

struct Poly {
    int n;
    Point p[MAXN];//p[n] = p[0]
    void init(Point *pp, int nn) {
        n = nn;
        for(int i = 0; i < n; ++i) p[i] = pp[i];
        p[n] = p[0];
    }
    double area() {
        if(n < 3) return 0;
        double s = p[0].y * (p[n - 1].x - p[1].x);
        for(int i = 1; i < n; ++i)
            s += p[i].y * (p[i - 1].x - p[i + 1].x);
        return s / 2;
    }
};

void Graham_scan(Point *p, int n, int *stk, int &top) {
    sort(p, p + n);
    top = 1;
    stk[0] = 0; stk[1] = 1;
    for(int i = 2; i < n; ++i) {
        while(top && cross(p[i], p[stk[top]], p[stk[top - 1]]) >= 0) --top;
        stk[++top] = i;
    }
    int len = top;
    stk[++top] = n - 2;
    for(int i = n - 3; i >= 0; --i) {
        while(top != len && cross(p[i], p[stk[top]], p[stk[top - 1]]) >= 0) --top;
        stk[++top] = i;
    }
}

/*******************************************************************************************/

Point p[MAXN];
Poly poly;
int stk[MAXN], top;
int n, T;

bool check() {
    poly.n = top;
    for(int i = 0; i < top; ++i) poly.p[i] = p[stk[i]];
    poly.p[poly.n] = poly.p[0];
    if(sgn(poly.area()) == 0) return false;
    for(int i = 0; i < poly.n; ++i) {
        bool flag = false;
        for(int j = 0; j < n; ++j) {
            if(p[j] == poly.p[i] || p[j] == poly.p[i + 1]) continue;
            if(isOnSeg(Seg(poly.p[i], poly.p[i + 1]), p[j])) {
                flag = true;
                break;
            }
        }
        if(!flag) return false;
    }
    return true;
}

int main() {
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(int i = 0; i < n; ++i) p[i].read();
        Graham_scan(p, n, stk, top);
        if(check()) puts("YES");
        else puts("NO");
    }
}