POJ 3525 Most Distant Point from the Sea 二分+半平面交
题目就是求多变形内部一点。 使得到任意边距离中的最小值最大。
那么我们想一下,可以发现其实求是看一个圆是否能放进这个多边形中。
那么我们就二分这个半径r,然后将多边形的每条边都往内退r距离。
求半平面交看是否存在解即可
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cstdlib> #include <cmath> #include <map> #include <sstream> #include <queue> #include <vector> #define MAXN 111111 #define MAXM 211111 #define PI acos(-1.0) #define eps 1e-8 #define INF 1000000001 using namespace std; int dblcmp(double d) { if (fabs(d) < eps) return 0; return d > eps ? 1 : -1; } struct point { double x, y; point(){} point(double _x, double _y): x(_x), y(_y){}; void input() { scanf("%lf%lf",&x, &y); } double dot(point p) { return x * p.x + y * p.y; } double distance(point p) { return hypot(x - p.x, y - p.y); } point sub(point p) { return point(x - p.x, y - p.y); } double det(point p) { return x * p.y - y * p.x; } bool operator == (point a)const { return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0; } bool operator < (point a)const { return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x; } }p[MAXN]; struct line { point a,b; line(){} line(point _a,point _b) { a=_a; b=_b; } bool parallel(line v) { return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0; } point crosspoint(line v) { double a1 = v.b.sub(v.a).det(a.sub(v.a)); double a2 = v.b.sub(v.a).det(b.sub(v.a)); return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1)); } bool operator == (line v)const { return (a == v.a) && (b == v.b); } }; struct halfplane:public line { double angle; halfplane(){} //表示向量 a->b逆时针(左侧)的半平面 halfplane(point _a, point _b) { a = _a; b = _b; } halfplane(line v) { a = v.a; b = v.b; } void calcangle() { angle = atan2(b.y - a.y, b.x - a.x); } bool operator <(const halfplane &b)const { return angle < b.angle; } }; struct polygon { int n; point p[MAXN]; line l[MAXN]; double area; void getline() { for (int i = 0; i < n; i++) { l[i] = line(p[i], p[(i + 1) % n]); } } void getarea() { area = 0; int a = 1, b = 2; while(b <= n - 1) { area += p[a].sub(p[0]).det(p[b].sub(p[0])); a++; b++; } area = fabs(area) / 2; } }convex; struct halfplanes { int n; halfplane hp[MAXN]; point p[MAXN]; int que[MAXN]; int st, ed; void push(halfplane tmp) { hp[n++] = tmp; } void unique() { int m = 1, i; for (i = 1; i < n;i++) { if (dblcmp(hp[i].angle - hp[i - 1].angle))hp[m++] = hp[i]; else if (dblcmp(hp[m - 1].b.sub(hp[m - 1].a).det(hp[i].a.sub(hp[m - 1].a)) > 0))hp[m - 1] = hp[i]; } n = m; } bool halfplaneinsert() { int i; for (i = 0; i < n; i++) hp[i].calcangle(); sort(hp, hp + n); unique(); que[st = 0] = 0; que[ed = 1] = 1; p[1] = hp[0].crosspoint(hp[1]); for (i = 2; i < n; i++) { while (st < ed && dblcmp((hp[i].b.sub(hp[i].a).det(p[ed].sub(hp[i].a)))) < 0) ed--; while (st < ed && dblcmp((hp[i].b.sub(hp[i].a).det(p[st + 1].sub(hp[i].a)))) < 0) st++; que[++ed] = i; if (hp[i].parallel(hp[que[ed - 1]])) return false; p[ed] = hp[i].crosspoint(hp[que[ed - 1]]); } while (st < ed && dblcmp(hp[que[st]].b.sub(hp[que[st]].a).det(p[ed].sub(hp[que[st]].a))) < 0) ed--; while (st < ed && dblcmp(hp[que[ed]].b.sub(hp[que[ed]].a).det(p[st + 1].sub(hp[que[ed]].a))) < 0) st++; if (st + 1 >= ed)return false; return true; } void getconvex(polygon &con) { p[st] = hp[que[st]].crosspoint(hp[que[ed]]); con.n = ed - st + 1; int j = st, i = 0; for (; j <= ed; i++, j++) { con.p[i] = p[j]; } } }h; int T; int n; line getmove(point a, point b, double mid) { double x = a.x - b.x; double y = a.y - b.y; double L = a.distance(b); point ta = point(mid * y / L + a.x, a.y - mid * x / L); point tb = point(mid * y / L + b.x, b.y - mid * x / L); return line(ta, tb); } bool check(double mid) { h.n = 0; for(int i = 0; i < n; i++) { line tmp = getmove(p[i], p[(i + 1) % n], mid); h.push(halfplane(tmp)); } return h.halfplaneinsert(); } int main() { int cas = 0; while(scanf("%d", &n) != EOF && n) { for(int i = 0; i < n; i++) p[i].input(); double low = 0, high = INF; for(int i = 0; i < 100; i++) { double mid = (low + high) / 2; if(check(mid)) low = mid; else high = mid; } printf("%.6f\n", low); } return 0; }
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