POJ 1271 Nice Milk （半平面交应用）

题目：传送门

题意：有一个凸多形面包，有一罐牛奶，牛奶高度为 h，认为牛奶的宽度无限长，现在你最多可以蘸 k 次牛奶， 问你蘸到牛奶的面包的面积最大是多少。

3 <= n <= 20, 0 <= k <= 8, 0 <= h <= 10

思路：dfs 枚举凸多边形蘸牛奶的 K 条边，将这些边向内缩进 h，求新的多边形的半平面交即是不能蘸到牛奶的面积，维护一个最小值，最后答案就是没有蘸牛奶时的面积减去这个最小值。

注意 k 可能大于 n，要判一下。

#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 = 22;
const double eps = 1e-6;
const double maxL = 1000.0;

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;
}

Point p[22];
Line q[22];

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[22], Q[22];

Line L[22], tmp_L[22];

void change(Point A, Point B, Point &C, Point &D, double p) { /// 求A,B两点向里移动d距离后的两点C,D
    double len = Length(B - A);
    double addx = (A.y - B.y) * p / len;
    double addy = (B.x - A.x) * p / len;
    C.x = A.x + addx; C.y = A.y + addy;
    D.x = B.x + addx; D.y = B.y + addy;
}

int n, k, h;
double mi = 100000000.0;

void dfs(int coun, int pos) {

    if(mi == 0.0) return ;
    if(n - pos < k - coun) return ;
    if(coun > k) return ;
    if(coun == k) {

        rep(i, 0, n - 1) tmp_L[i] = L[i];

        int cnt = HPI(tmp_L, n, Q);

        double res = 0.0;

        rep(i, 1, cnt - 2) res += Cross(Q[i] - Q[0], Q[i + 1] - Q[0]);
        if(res < 0) res = -res;

        mi = min(mi, res);
        return ;
    }

    if(pos >= n) return ;

    Point A, B;
    change(P[pos], P[(pos + 1) % n], A, B, h);
    L[pos] = Line(A, B - A);
    dfs(coun + 1, pos + 1);
    L[pos] = Line(P[pos], P[(pos + 1) % n] - P[pos]);
    dfs(coun, pos + 1);
}

int main() {

    while(~scanf("%d %d %d", &n, &k, &h)) {
        if(n == 0 && k == 0 && h == 0) return 0;
        if(k > n) k = n;
        rep(i, 0, n - 1) scanf("%lf %lf", &P[i].x, &P[i].y);

        if(k == 0 || h == 0) {
            puts("0.00"); continue;
        }

        rep(i, 0, n - 1) {
            L[i] = Line(P[i], P[(i + 1) % n] - P[i]);
            tmp_L[i] = L[i];
        }
        double ans = 0;

        int cnt = HPI(tmp_L, n, Q);

        rep(i, 1, cnt - 2) ans += Cross(Q[i] - Q[0], Q[i + 1] - Q[0]);
        if(ans < 0) ans = -ans;

        mi = 100000000.0;

        Point A, B;
        change(P[0], P[1], A, B, h);
        L[0] = Line(A, B - A);
        dfs(1, 1);
        L[0] = Line(P[0], P[1] - P[0]);
        dfs(0, 1);

        printf("%.2f
", (ans - mi) / 2.0);

    }

    return 0;
}