POJ 2540 Hotter Colder (半平面交应用 + 向量旋转)
题目:传送门
题意:在一个以 (0, 0) 为左下角,以(10, 10) 为右上角的正方形中。正方形的某个点藏着东西,你会猜测这个东西在哪里。起初你在点 (0,0),随后给出最多 50 个点,表示的是你猜测的点,然后还会输入一个字符串,若输入的是Hotter,表示你当前所在的点离那个东西的距离比你之前所在的那个点离那个东西的距离小,也就是你离那个东西更近了,Colder 则是更远了,Same 则是相等。
思路:
将新输入的点和你之前所在的点连线,它们的中垂线就是离他们一样近的点,那可以通过输入的字符串,来判断可能的区域在线的左边还是右边,那就是求许多不等式围成的面积,用半平面交即可解得,如果输入的是 same 那面积一定是 0.
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <map> #include <vector> #include <set> #include <string> #include <math.h> #define LL long long #define mem(i, j) memset(i, j, sizeof(i)) #define rep(i, j, k) for(int i = j; i <= k; i++) #define dep(i, j, k) for(int i = k; i >= j; i--) #define pb push_back #define make make_pair #define INF INT_MAX #define inf LLONG_MAX #define PI acos(-1) #define fir first #define sec second using namespace std; const int N = 15000; const double eps = 1e-8; const double maxL = 10.0; struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } }; int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } Point operator + (Point A, Point B) { return Point(A.x + B.x, A.y + B.y); } Point operator - (Point A, Point B) { return Point(A.x - B.x, A.y - B.y); } Point operator * (Point A, double p) { return Point(A.x * p, A.y * p); } Point operator / (Point A, double p) { return Point(A.x / p, A.y / p); } double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; } double Dot(Point A, Point B) { return A.x * B.x + A.y * B.y; } double Length(Point A) { return sqrt(Dot(A, A)); } Point Rotate(Point A, double rad) { /// 向量逆时针旋转 rad (弧度) return Point(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); } /* 有向直线,它的左边就是对应的半平面 */ struct Line { Point p; /// 直线任意一点 Point v; /// 方向向量 double ang; /// 极角,即从x正半轴旋转到向量v所需要的角(弧度) Line() { } Line(Point p, Point v) : p(p), v(v) { ang = atan2(v.y, v.x); } bool operator < (const Line& L) const { return ang < L.ang; } }; /* 点p在有向直线L的左边 */ bool OnLeft(Line L, Point p) { return dcmp(Cross(L.v, p - L.p)) > 0; } /* 二直线交点,假设交点唯一存在。*/ Point GLI(Line a, Line b) { Point u = a.p - b.p; double t = Cross(b.v, u) / Cross(a.v, b.v); return a.p + a.v * t; } int HPI(Line* L, int n, Point* Q) { sort(L, L + n); /// 极角排序 int st, ed; /// 双端队列的第一个元素和最后一个元素的下标 Point *p = new Point[n]; /// p[i]为q[i]和q[i+1]的交点 Line *q = new Line[n]; /// 双端队列 q[st = ed = 0] = L[0]; rep(i, 1, n - 1) { while(st < ed && !OnLeft(L[i], p[ed - 1])) ed--; while(st < ed && !OnLeft(L[i], p[st])) st++; q[++ed] = L[i]; /// 平行取内测那条 if(fabs(Cross(q[ed].v, q[ed - 1].v)) < eps) { ed--; if(OnLeft(q[ed], L[i].p)) q[ed] = L[i]; } if(st < ed) p[ed - 1] = GLI(q[ed - 1], q[ed]); } while(st < ed && !OnLeft(q[st], p[ed - 1])) ed--; if(ed - st <= 1) return 0; p[ed] = GLI(q[ed], q[st]); int m = 0; rep(i, st, ed) Q[m++] = p[i]; return m; } Point P[N], Q[N]; Line L[N]; char s[10]; void solve() { Point now_Point, tmpa, tmpb; Point pre_Point = Point(0.0, 0.0); int cnt = 4; ///给半平面加一个框,这样可以使解x,y都大于0,也可以避免所有半平面交起来后为不为凸多边形,而是一个敞开的区域 ///如果题目输入的不是一个多边形,而是本题这种输入若干不等式组的情况,这样的限定就是必须的,不然有bug,例如,两条线是平行的(但是极角不同), ///极角排序后又挨在一起, 那么就可能求它们的交点,就容易出错 tmpa.x = 0; tmpa.y = 0; tmpb.x = maxL; tmpb.y = 0; L[0] = Line(tmpa, tmpb - tmpa); tmpa = tmpb; tmpb.x = maxL; tmpb.y = maxL; L[1] = Line(tmpa, tmpb - tmpa); tmpa = tmpb; tmpb.x = 0; L[2] = Line(tmpa, tmpb - tmpa); tmpa = tmpb; tmpb.y = 0; L[3] = Line(tmpa, tmpb - tmpa); bool flag = 0; while(scanf("%lf %lf %s", &now_Point.x, &now_Point.y, s) != EOF) { if(flag) { puts("0.00"); continue; } Point v; if(s[0] == 'C') { v = now_Point - pre_Point; } else if(s[0] == 'H') { v = pre_Point - now_Point; } else if(s[0] == 'S') { flag = 1; puts("0.00"); continue; } v = Rotate(v, PI / 2.0); Point tmp = (now_Point + pre_Point) / 2; L[cnt++] = Line(tmp, v); int n = HPI(L, cnt, Q); double ans = 0.0; rep(i, 1, n - 2) { ans += Cross(Q[i] - Q[0], Q[i + 1] - Q[0]); } if(ans < 0) ans = -ans; ans /= 2.0; printf("%.2f\n", ans); pre_Point = now_Point; } } int main() { // int _; scanf("%d", &_); // while(_--) solve(); solve(); return 0; }
一步一步,永不停息