CRB and Farm HDU

CRB and Farm

HDU - 5408

题意：给一个ｎ个点的大凸包，再给ｋ个内部点，问能否选择不超过２ｋ个大凸包上的点使得所选点围成的凸包完全包含内部的ｋ个点．

首先明确一定是存在的

可以先对内部的ｋ个点求一个凸包，然后选一个凸包内的点Ａ，求Ａ与凸包上的各点形成的射线与大凸包的交点．则交点所在线段的两端就是所选点．

 

关键在于判断射线与线段相交．

//判断射线a1--b1是否与线段a2--b2相交
bool Intersection(Point a1, Point b1, Point a2, Point b2) { 
    double c1 = Cross(b1-a1, a2-a1), c2 = Cross(b1-a2, b2-a2);
    double c3 = Cross(b1-b2, a1-b2);
    return dcmp(c1)<0 && dcmp(c2)<0 && dcmp(c3)<0 ||
           dcmp(c1)>0 && dcmp(c2)>0 && dcmp(c3)>0;

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 2e5+10;

struct Point {
    double x,y;
    Point (double x = 0, double y = 0) : x(x), y(y) {}
};
typedef Point Vector;
Vector operator + (Vector a, Vector b) {
    return Vector (a.x + b.x, a.y + b.y);
}
Vector operator * (Vector a, double s) {
    return Vector (a.x * s, a.y * s);
}
Vector operator / (Vector a, double p) {
    return Vector (a.x / p, a.y / p);
}
Vector operator - (Point a, Point b) {
    return Vector (a.x - b.x, a.y - b.y);
}
bool operator < (Point a, Point b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
int dcmp (double x) {
    if(fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}
bool operator == (const Point &a, const Point &b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector a, Vector b) {
    return a.x * b.x + a.y * b.y;
}
double Length (Vector a) {
    return sqrt(Dot(a, a));
}
double Angle (Vector a, Vector b) {
    return acos(Dot(a, b) / Length(a) / Length(b));
}
double Cross (Vector a, Vector b) {
    return a.x * b.y - a.y * b.x;
}
//判断射线a1--b1是否与线段a2--b2相交
bool Intersection(Point a1, Point b1, Point a2, Point b2) { 
    double c1 = Cross(b1-a1, a2-a1), c2 = Cross(b1-a2, b2-a2);
    double c3 = Cross(b1-b2, a1-b2);
    return dcmp(c1)<0 && dcmp(c2)<0 && dcmp(c3)<0 ||
           dcmp(c1)>0 && dcmp(c2)>0 && dcmp(c3)>0;
}
//凸包
int ConvexHull(Point *p, int n, Point *ch) {
    int m = 0;
    sort(p, p+n);
    for(int i = 0; i < n; i++) {
        while(m > 1 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;  //去掉在边上的点
        ch[m++] = p[i];
    }
    int k = m;
    for(int i = n-2; i >= 0; i--) {
        while(m > k && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;  //
        ch[m++] = p[i];
    }
    if(m > 1) m--;
    return m;
}

Point p1[maxn], p2[maxn], ch[maxn];

int main() {
    int t;
    //freopen("in.txt", "r", stdin);
    scanf("%d", &t);
    while(t--){
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            scanf("%lf %lf", &p1[i].x, &p1[i].y);
        }
        p1[n] = p1[0];
        int k;
        scanf("%d", &k);
        for(int i = 0; i < k; i++){
            scanf("%lf %lf", &p2[i].x, &p2[i].y);
        }
        int cnt = ConvexHull(p2, k, ch);
        Point o = ch[0];
        vector<int> v;
        for(int i = 1; i < cnt; i++) {
            o = (o+ch[i])/2;
        }
        int id = 0;
        for(int i = 0; i < cnt; i++){
            while(!Intersection(o, ch[i], p1[id], p1[id+1])) id=(id+1)%n;
            v.push_back(id);
            v.push_back((id+1)%n);
        }
        sort(v.begin(), v.end());
        cnt = unique(v.begin(), v.end()) - v.begin();
        puts("Yes");
        printf("%d
", cnt);
        for(int i = 0; i < cnt; i++) printf("%d%c", v[i]+1, i==cnt-1?'
':' '); 
    }
}