[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;
}