[CQOI2006]凸多边形（半平面相交）

嘟嘟嘟




本来我要写feng shui这道题的。然后网上都说什么半平面相交，于是我还得现学这个东西，就来刷这道模板题了。




所谓的半平面相交和高中数学的分数规划特别像。比如这道题，把每一条边看成一条有向直线，则合法的范围都是直线的右半部分，最后求交集。大概是每一次都取一半，所以就叫半平面相交吧。




\(O(n ^ 2)\)的做法很简单，我也只会\(O(n ^ 2)\)的。枚举每一条边，然后用这条边去切当前算出来的图形。

具体怎么切？一句话就是把这条直线左边的点全部扔掉。

放个伪代码就明白了：

for 每条边ai ai+1

	if (ai在AB右边)

		把ai加入答案

		if (ai+1在AB左边) 把交点加入答案

	else if(ai+1在AB右边) 把交点加入答案

至于判断左右，用叉积求又向面积就行了。

#include<cstdio>

#include<iostream>

#include<cmath>

#include<algorithm>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<vector>

#include<stack>

#include<queue>

using namespace std;

#define enter puts("")

#define space putchar(' ')

#define Mem(a, x) memset(a, x, sizeof(a))

#define rg register

typedef long long ll;

typedef double db;

const int INF = 0x3f3f3f3f;

const db eps = 1e-8;

const int maxn = 5e5 + 5;

inline ll read()

{

	ll ans = 0;

	char ch = getchar(), last = ' ';

	while(!isdigit(ch)) {last = ch; ch = getchar();}

	while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}

	if(last == '-') ans = -ans;

	return ans;

}

inline void write(ll x)

{

	if(x < 0) x = -x, putchar('-');

	if(x >= 10) write(x / 10);

	putchar(x % 10 + '0');

}

int n, m, cnt = 0;

struct Point

{

	db x, y;

	Point operator - (const Point& oth)const

	{

		return (Point){x - oth.x, y - oth.y};

	}

	db operator * (const Point& oth)const

	{

		return x * oth.y - oth.x * y;

	}

	Point operator * (const db& d)const

	{

		return (Point){x * d, y * d};

	}

}p[maxn], a[maxn];

int tot = 0;

Point b[maxn];

db cross(Point A, Point B, Point C)

{

	return (B - A) * (C - A);

}

void addCross(Point A, Point B, Point C, Point D)

{

	db s1 = (C - A) * (D - A), s2 = (D - B) * (C - B);

	b[++tot] = A - (A - B) * (s1 / (s1 + s2));

}

void cut(Point A, Point B)

{

	tot = 0;

	a[cnt + 1] = a[1];

	for(int i = 1; i <= cnt; ++i)

	{

		if(cross(A, B, a[i]) >= 0)

		{

			b[++tot] = a[i];

			if(cross(A, B, a[i + 1]) < 0) addCross(A, B, a[i], a[i + 1]);

		}

		else if(cross(A, B, a[i + 1]) > 0) addCross(A, B, a[i], a[i + 1]);

	}

	for(int i = 1; i <= tot; ++i) a[i] = b[i];

	cnt = tot;

}

int main()

{

	n = read(); m = read();

	for(int i = 1; i <= m; ++i) a[i].x = read(), a[i].y = read();

	cnt = m; n--;

	while(n--)

	{

		m = read();

		for(int i = 1; i <= m; ++i) p[i].x = read(), p[i].y = read();

		p[m + 1] = p[1];

		for(int i = 1; i <= m; ++i) cut(p[i], p[i + 1]);

	}

	a[cnt + 1] = a[1];

	db ans = 0;

	for(int i = 1; i <= cnt; ++i) ans += a[i] * a[i + 1];

	printf("%.3lf\n", ans / 2);

	return 0;

}