AK F.*ing leetcode 流浪计划之半平面求交

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本期话题：半平面求交

背景知识

学习资料

视频讲解
https://www.bilibili.com/video/BV1jL411C7Ct/?spm_id_from=333.1007.top_right_bar_window_history.content.click&vd_source=fb27f95f25902a2cc94d4d8e49f5f777

文本资料
https://oi-wiki.org//geometry/half-plane/

基本问题转化

在很多题目中，给定的线段是没有方向的。此时，我们需要先把所有的线段都转化成点加向量的方式。使得向量的左边为有效区域。这样就可以使用模板求解了。

要注意的问题

1. 主要的问题是浮点型的判断大小问题。在排序和判断点与线的关系时都用到浮点型判断。有些题型会卡精度，能用整数判断尽量不要使用浮点判断。
2. atan2计算比较耗时，可以事先保存。

代码模板

求多边形的核

题目链接：https://vjudge.net/problem/UVA-1571

多边形的核就是取核区域内任意一点，站在该可以观察到多边形内任意一点。

利用半平面求交可以得到多边形的核

#include<stdio.h>
#include<cmath>
#include <algorithm>
#include <vector>
#include <list>
#include <cstring>
#include <set>using namespace std;
const double EPS = 1e-14;const int N = 2e6 + 10;int cmp(double d) {if (abs(d) < EPS)return 0;if (d > 0)return 1;return -1;
}class Point {
public:double x, y;int id;Point() {}Point(double a, double b) :x(a), y(b) {}Point(const Point& p) :x(p.x), y(p.y), id(p.id) {}void in() {scanf("%lf %lf", &x, &y);}void out() {printf("%.16f %.16f\n", x, y);}double dis() {return sqrt(x * x + y * y);}double dis2() {return x * x + y * y;}Point operator -() const {return Point(-x, -y);}Point operator -(const Point& p) const {return Point(x - p.x, y - p.y);}Point operator +(const Point& p) const {return Point(x + p.x, y + p.y);}Point operator *(double d)const {return Point(x * d, y * d);}Point operator /(double d)const {return Point(x / d, y / d);}void operator -=(Point& p) {x -= p.x;y -= p.y;}void operator +=(Point& p) {x += p.x;y += p.y;}void operator *=(double d) {x *= d;y *= d;}void operator /=(double d) {this ->operator*= (1 / d);}bool operator<(const Point& a) const {return x < a.x || (abs(x - a.x) < EPS && y < a.y);}bool operator==(const Point& a) const {return abs(x - a.x) < EPS && abs(y - a.y) < EPS;}
};// 向量操作double cross(const Point& a, const Point& b) {return a.x * b.y - a.y * b.x;
}double dot(const Point& a, const Point& b) {return a.x * b.x + a.y * b.y;
}class Line {
public:Point front, tail;double ang;int u, v;Line() {}Line(const Point& a, const Point& b) :front(a), tail(b) {ang = atan2(front.y - tail.y, front.x - tail.x);}
};int cmp(const Line& a, const Line& b) {//if (a.u == b.u && a.v == b.v)return 0;return cmp(a.ang - b.ang);}// 点在直线哪一边>0 左边，<0边
double SideJudge(const Line& a, const Point& b) {//return cmp(cross(a.front - a.tail, b - a.tail));return cross(a.front - a.tail, b - a.tail);
}int LineSort(const Line& a, const Line& b) {int c = cmp(a, b);if (c)return c < 0;return	cross(b.front - b.tail, a.front - b.tail) > 0;
}/*
点p 到 p+r 表示线段1
点q 到 q+s 表示线段2
线段1 上1点用 p' = p+t*r (0<=t<=1)
线段2 上1点用 q' = q+u*s (0<=u<=1)
让两式相等求交点 p+t*r = q+u*s
两边都叉乘s
(p+t*r)Xs = (q+u*s)Xs
pXs + t*rXs = qXs
t = (q-p)Xs/(rXs)
同理，
u = (p-q)Xr/(sXr) -> u = (q-p)Xr/(rXs)以下分4种情况：
1. 共线，sXr==0 && (q-p)Xr==0, 计算 (q-p)在r上的投影在r长度上的占比t0，
计算(q+s-p)在r上的投影在r长度上的占比t1，查看[t0, t1]是否与范围[0，1]有交集。
如果t0>t1, 则比较[t1, t0]是否与范围[0，1]有交集。
t0 = (q-p)*r/(r*r)
t1 = (q+s-p)*r/(r*r) = t0 + s · r / (r · r)
2. 平行sXr==0 && (q-p)Xr!=0
3. 0<=u<=1 && 0<=t<=1 有交点
4. 其他u, t不在0到范围内，没有交点。
*/
pair<double, double> intersection(const Point& q, const Point& s, const Point& p, const Point& r, bool &oneline) {// 计算 (q-p)Xrauto qpr = cross(q - p, r);auto qps = cross(q - p, s);auto rXs = cross(r, s);if (cmp(rXs) == 0) {oneline = true;return { -1, -1 }; // 平行或共线}// 求解t, u// t = (q-p)Xs/(rXs)auto t = qps / rXs;// u = (q-p)Xr/(rXs)auto u = qpr / rXs;return { u, t };
}Point LineCross(const Line& a, const Line& b, bool &f) {Point dira = a.front - a.tail;Point dirb = b.front - b.tail;bool oneline=false;auto p = intersection(a.tail, dira, b.tail, dirb, oneline);if (oneline)f = false;return a.tail + dira * p.first;
}class HalfPlane {
public:vector<Line> lines;vector<int> q;vector<Point> t;int len;HalfPlane() {lines.resize(N);q.resize(N);t.resize(N);}void reset() {len = 0;}void addLine(const Line& a) {lines[len++] = a;}bool run() {sort(lines.begin(), lines.begin() + len, LineSort);int l = -1, r = 0;q[0] = 0;for (int i = 1; i < len; ++i) {if (cmp(lines[i], lines[i - 1]) == 0)continue;while (r - l > 1 && SideJudge(lines[i], t[r]) < 0)r--;while (r - l > 1 && SideJudge(lines[i], t[l + 2]) < 0)l++;q[++r] = i;bool f=true;t[r] = LineCross(lines[q[r]], lines[q[r - 1]], f);}while (r - l > 1 && SideJudge(lines[q[l + 1]], t[r]) < 0)r--;//if (r - l > 1) {//	bool f = true;//	t[r + 1] = LineCross(lines[q[l + 1]], lines[q[r]], f);//	r++;//	if (!f)r -= 2;//} 统计交点//l++;//vector<Point> ans(r - l);//for (int i = 0; i < ans.size(); ++i) {//	ans[i] = t[i + l + 1];//}return r-l>2;}
};Point oiPs[N * 2];
pair<int, int> ori[N * 2];
HalfPlane hp;int bigDevid(int a, int b) {for (int i = max(abs(a), abs(b)); i >= 1; i--) {if (a % i == 0 && b % i == 0)return i;}return 1;
}void  solve() {int n, m = 1;//FILE* fp = fopen("ans.txt", "w");while (scanf("%d", &n) != EOF && n) {int a, b;for (int i = 0; i < n; ++i) {scanf("%d %d", &a, &b);oiPs[i] = Point(a, b);ori[i] = { a,b };}oiPs[n] = oiPs[0];ori[n] = ori[0];hp.reset();for (int i = 0; i < n; ++i) {hp.addLine(Line(oiPs[i+1], oiPs[i]));hp.lines[i].u = ori[i+1].first - ori[i].first;hp.lines[i].v = ori[i+1].second - ori[i].second;int bd = bigDevid(hp.lines[i].u, hp.lines[i].v);hp.lines[i].u /= bd;hp.lines[i].v /= bd;}auto ps = hp.run();if (ps)puts("1");else puts("0");m++;}
}int main() {solve();return 0;}/*
4
0 0
0 1
1 1
1 0
8
0 0
3 0
4 3
2 2
3 4
4 4
4 5
0 5
08
0 0
0 1
1 1
1 2
0 2
0 3
3 3
3 0
*/

练习一

链接：https://www.luogu.com.cn/problem/P4196

求多个凸多边形的交面积。

对每条边进行半平面求交，再利用三角形求多边形面积。

#include<stdio.h>
#include<cmath>
#include <algorithm>
#include <vector>
#include <list>
#include <cstring>
#include <set>using namespace std;
const double EPS = 1e-14;const int N = 2e6 + 10;int cmp(double d) {if (abs(d) < EPS)return 0;if (d > 0)return 1;return -1;
}class Point {
public:double x, y;int id;Point() {}Point(double a, double b) :x(a), y(b) {}Point(const Point& p) :x(p.x), y(p.y), id(p.id) {}void in() {scanf("%lf %lf", &x, &y);}void out() {printf("%.16f %.16f\n", x, y);}double dis() {return sqrt(x * x + y * y);}double dis2() {return x * x + y * y;}Point operator -() const {return Point(-x, -y);}Point operator -(const Point& p) const {return Point(x - p.x, y - p.y);}Point operator +(const Point& p) const {return Point(x + p.x, y + p.y);}Point operator *(double d)const {return Point(x * d, y * d);}Point operator /(double d)const {return Point(x / d, y / d);}void operator -=(Point& p) {x -= p.x;y -= p.y;}void operator +=(Point& p) {x += p.x;y += p.y;}void operator *=(double d) {x *= d;y *= d;}void operator /=(double d) {this ->operator*= (1 / d);}bool operator<(const Point& a) const {return x < a.x || (abs(x - a.x) < EPS && y < a.y);}bool operator==(const Point& a) const {return abs(x - a.x) < EPS && abs(y - a.y) < EPS;}
};// 向量操作double cross(const Point& a, const Point& b) {return a.x * b.y - a.y * b.x;
}double dot(const Point& a, const Point& b) {return a.x * b.x + a.y * b.y;
}class Line {
public:Point front, tail;double ang;int u, v;Line() {}Line(const Point& a, const Point& b) :front(a), tail(b) {ang = atan2(front.y - tail.y, front.x - tail.x);}
};int cmp(const Line& a, const Line& b) {//if (a.u == b.u && a.v == b.v)return 0;return cmp(a.ang - b.ang);}// 点在直线哪一边>0 左边，<0边
double SideJudge(const Line& a, const Point& b) {//return cmp(cross(a.front - a.tail, b - a.tail));return cross(a.front - a.tail, b - a.tail);
}int LineSort(const Line& a, const Line& b) {int c = cmp(a, b);if (c)return c < 0;return	cross(b.front - b.tail, a.front - b.tail) > 0;
}/*
点p 到 p+r 表示线段1
点q 到 q+s 表示线段2
线段1 上1点用 p' = p+t*r (0<=t<=1)
线段2 上1点用 q' = q+u*s (0<=u<=1)
让两式相等求交点 p+t*r = q+u*s
两边都叉乘s
(p+t*r)Xs = (q+u*s)Xs
pXs + t*rXs = qXs
t = (q-p)Xs/(rXs)
同理，
u = (p-q)Xr/(sXr) -> u = (q-p)Xr/(rXs)以下分4种情况：
1. 共线，sXr==0 && (q-p)Xr==0, 计算 (q-p)在r上的投影在r长度上的占比t0，
计算(q+s-p)在r上的投影在r长度上的占比t1，查看[t0, t1]是否与范围[0，1]有交集。
如果t0>t1, 则比较[t1, t0]是否与范围[0，1]有交集。
t0 = (q-p)*r/(r*r)
t1 = (q+s-p)*r/(r*r) = t0 + s · r / (r · r)
2. 平行sXr==0 && (q-p)Xr!=0
3. 0<=u<=1 && 0<=t<=1 有交点
4. 其他u, t不在0到范围内，没有交点。
*/
pair<double, double> intersection(const Point& q, const Point& s, const Point& p, const Point& r, bool &oneline) {// 计算 (q-p)Xrauto qpr = cross(q - p, r);auto qps = cross(q - p, s);auto rXs = cross(r, s);if (cmp(rXs) == 0) {oneline = true;return { -1, -1 }; // 平行或共线}// 求解t, u// t = (q-p)Xs/(rXs)auto t = qps / rXs;// u = (q-p)Xr/(rXs)auto u = qpr / rXs;return { u, t };
}Point LineCross(const Line& a, const Line& b, bool &f) {Point dira = a.front - a.tail;Point dirb = b.front - b.tail;bool oneline=false;auto p = intersection(a.tail, dira, b.tail, dirb, oneline);if (oneline)f = false;return a.tail + dira * p.first;
}class HalfPlane {
public:vector<Line> lines;void addLine(const Line& a) {lines.push_back(a);}vector<Point> run() {sort(lines.begin(), lines.end(), LineSort);vector<int> q(lines.size() + 10);vector<Point> t(lines.size() + 10);int l = -1, r = 0;q[0] = 0;for (int i = 1; i < lines.size(); ++i) {if (cmp(lines[i], lines[i - 1]) == 0)continue;while (r - l > 1 && SideJudge(lines[i], t[r]) < 0)r--;while (r - l > 1 && SideJudge(lines[i], t[l + 2]) < 0)l++;q[++r] = i;bool f = true;t[r] = LineCross(lines[q[r]], lines[q[r - 1]], f);}while (r - l > 1 && SideJudge(lines[q[l + 1]], t[r]) < 0)r--;if (r - l > 1) {bool f = true;t[r + 1] = LineCross(lines[q[l + 1]], lines[q[r]], f);r++;}// 统计交点l++;vector<Point> ans(r - l);for (int i = 0; i < ans.size(); ++i) {ans[i] = t[i + l + 1];}return ans;}
};Point oiPs[N];void  solve() {int n, m;scanf("%d", &n);HalfPlane hp;int a, b;while (n--) {scanf("%d", &m);for (int i = 0; i < m; ++i) {scanf("%d%d", &a, &b);oiPs[i].x = a;oiPs[i].y = b;}oiPs[m] = oiPs[0];for (int i = 0; i < m; ++i) {hp.addLine(Line(oiPs[i + 1], oiPs[i]));}}auto keyPoints = hp.run();double ans = 0;for (int i = 2; i < keyPoints.size(); ++i) {ans += cross(keyPoints[i - 1] - keyPoints[0], keyPoints[i] - keyPoints[0]);}printf("%.3f\n", ans / 2);
}int main() {solve();return 0;}/*
3
3
-1 2
-2 1
-1 13
1 1
2 1
1 23
1 1
3 0
2 2*/

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