POJ 3384 Feng Shui (半平面交)

　　风水。。。是一门艺术。。。我在想要不要看看《周易》《葬经》什么的，等老了还能出去给人算算命，看看阴宅阳宅什么的，混口饭吃。嘿嘿，扯远了。。。

　　题意:给一个凸多边形，然后在里面放两个半径为r的圆。问怎么放能使覆盖的面积最大(spj)。

　　思路：把多边形的每一条边向“内”移r的距离，交得一个新多边形。在多边形上找两个尽量圆的点，放上圆心。

关于怎么向内移动r的距离，见下图

所以 point[i] 和 point[i+1] x轴上移动的距离就是r*cos(th), y轴上移动距离就是r*sin(th)；

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int inf = ~0u>>2;

using namespace std;

const int maxn = 1010;

struct Point {
    double x;
    double y;
    Point() {}
    Point(double a, double b): x(a), y(b) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
};

Point point[maxn], p[maxn], q[maxn];    //读入的多边形的顶点（顺时针）、p为存放最终切割得到的多边形顶点的数组、暂存核的顶点
int cCnt, n;    //此时cCnt为最终切割得到的多边形的顶点数、暂存顶点个数

double r;

inline int dbcmp(double x) {    //精度问题
    if(x > eps) return 1;
    else if(x < -eps)   return -1;
    return 0;
}

inline void getline(Point x, Point y, double& a, double& b, double& c) {    //点X，Y确定一条直线
    a = y.y - x.y;
    b = x.x - y.x;
    c = y.x*x.y - x.x*y.y;
}

inline Point intersect(Point x, Point y, double a, double b, double c) {    ////求x、y形成的直线与已知直线a、b、c、的交点
    double u = fabs(a*x.x + b*x.y + c);
    double v = fabs(a*y.x + b*y.y + c);
    return Point((x.x*v + y.x*u)/(u + v), (x.y*v + y.y*u)/(u + v));
}

inline void cut(double a, double b, double c) {        //如上图所示，切割
    int cur = 0, i;
    for(i = 1; i <= cCnt; ++i) {
        if(dbcmp(a*p[i].x + b*p[i].y + c) >= 0)  q[++cur] = p[i];    // c由于精度问题，可能会偏小，所以有些点本应在右侧而没在
        else {
            if(dbcmp(a*p[i-1].x + b*p[i-1].y + c) > 0)    //如果p[i-1]在直线的右侧的话，
                //则将p[i],p[i-1]形成的直线与已知直线的交点作为核的一个顶点(这样的话，由于精度的问题，核的面积可能会有所减少)
                q[++cur] = intersect(p[i], p[i-1], a, b, c);
            if(dbcmp(a*p[i+1].x + b*p[i+1].y + c) > 0)
                q[++cur] = intersect(p[i], p[i+1], a, b, c);
        }
    }
    for(i = 1; i <= cur; ++i)   p[i] = q[i];
    p[cur+1] = q[1]; p[0] = p[cur];
    cCnt = cur;
}

double ddis(Point a, Point b) {
    return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}

void solve() {    //注意：默认点是顺时针，如果题目不是顺时针，规整化方向
    int i, j;
    Point pa, pb, pt;
    double Cos, Sin;
    double a, b, c;

    for(i = 1; i <= n; ++i) {
        a = point[i].x - point[i+1].x;    //point[]不是p[]....
        b = point[i+1].y - point[i].y;

        Cos = b/sqrt(a*a + b*b);
        Sin = a/sqrt(a*a + b*b);

        pt = Point(r*Cos, r*Sin);
        pa = Point(point[i].x + pt.x, point[i].y + pt.y);
        pb = Point(point[i+1].x + pt.x, point[i+1].y + pt.y);

        getline(pa, pb, a, b, c);
        cut(a, b, c);
    }

    double tmp, maxd = 0;
    int a1 = 0, a2 = 0;
    for(i = 1; i <= cCnt; ++i) {
        for(j = i + 1; j <= cCnt; ++j) {
            tmp = ddis(p[i], p[j]);
            //printf("%d %d %f\n", i, j, tmp);
            if(dbcmp(tmp - maxd) > 0) {
                maxd = tmp;
                a1 = i;
                a2 = j;
            }
        }
    }
    printf("%.4f %.4f %.4f %.4f\n", p[a1].x, p[a1].y, p[a2].x, p[a2].y);
}

void init() {
    int i;
    FOR(i, 1, n)    point[i].input();
    point[n+1] = point[1];
    //初始化p[], cCnt
    FOR(i, 1, n)    p[i] = point[i];
    p[n+1] = p[1]; p[0] = p[n];
    cCnt = n;
}

int main() {
    //freopen("data.in", "r", stdin);

    while(~scanf("%d%lf", &n, &r)) {
        init();
        solve();
    }
    return 0;
}