hdu6755 Mow

半平面交+数组模拟双端队列

人生第一次代码过两百行啊...加油加油

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<iomanip>

#define ld long double
using namespace std;

const int maxn = 1e3;
const ld pi = acosl(-);
const ld eps = 1e-;//控制精度 

int n;
int head, tail;
ld r, cost1, cost2;

typedef struct Grid
{//棋盘
    ld x,y;
    Grid(double a = , double b = ){x = a, y = b;}//构造函数，结构体对象创建时执行
} point , Vector ;//point:类型为Grid的数据类型；Vector与point结构相同，数据类型名称不同 

typedef struct Net
{
    point a,b;
    Net() {}//重载构造函数，目的是以形如line x;的方式定义直线，避免定义时确定参数的复杂性
    Net(point i, point j){a = i, b = j;}
} line ;

point node[maxn], p[maxn];
line l[maxn], L[maxn], q[maxn];

Vector operator - (point a, point b) // V需大写，避免与vector混淆
{
    return Vector(b.x - a.x, b.y - a.y);
}//矢量定义 

double operator * (Vector a, Vector b)
{
    return (a.x * b.y - a.y * b.x);
}//叉乘 

ld getangle(Vector t)
{
    return atan2(t.y, t.x);//返回一个(-pi,pi]的角度
}//获得某个矢量的极角 

ld getangle(line t)
{
    return atan2(t.b.y - t.a.y, t.b.x - t.a.x);
}//获得某条直线的极角 

bool cmp(line a, line b)
{
    Vector v1 = (a.b - a.a), v2 = (b.b - b.a);
    ld A = getangle(v1), B = getangle(v2);
    if(fabs(A - B) < eps){
        return (v1 * (b.b - a.a)  >= );//靠左边的丢在后面，方便双端队列去重
    }
    return A < B;
}//决定根据极角排序的直线的优先级 

bool check()
{
    ld ans = ;
    for(int i =  ; i < n- ; i++){
        ans += (node[i] - node[]) * (node[i+] - node[]);
    }
    if(ans < )    return false;
    return true;
}//检查point输入顺序进行调整 true为逆时针 

point getnode(line a, line b)//求两直线(线段)交点，一般式三参数暴力求解
{
    ld a1 = a.b.y - a.a.y, b1 = a.a.x - a.b.x, c1 = a.b.x * a.a.y - a.a.x * a.b.y;
    ld a2 = b.b.y - b.a.y, b2 = b.a.x - b.b.x, c2 = b.b.x * b.a.y - b.a.x * b.b.y;
    ld D = a1 * b2 - a2 * b1;
    //Dx/D,Dy/D,注意c1、c2移至右侧变号
    return point((c2 * b1 - b2 * c1) / D, (c1 * a2 - a1 * c2) / D);
}

bool on_right(line a, line b, line c)//观察a、b交点与c的关系
{
    point n = getnode(a, b);
    if((c.b - c.a) * (n - c.a) < ){//交点在右侧/下方
         return true;
    }
    return false;
}

line narrow_line(line x)//实现直线平移，勾画出圆心轨迹
{
    ld dx = x.b.x - x.a.x, dy = x.b.y - x.a.y;
    Vector turnv;//旋转矢量
    turnv.x = -dy, turnv.y = dx;

    ld m = sqrtl(dx * dx + dy * dy);
    turnv.x = turnv.x / m * r;
    turnv.y = turnv.y / m * r;
    //求对应坐标
    x.b.x += turnv.x;
    x.b.y += turnv.y;
    x.a.x += turnv.x;
    x.a.y += turnv.y;
    return x;
}

void narrow_polygon()//模拟圆心轨迹
{
    for(int i =  ; i <= n- ; i++){
        L[i] = narrow_line(l[i]);
    }
}

bool HPI()
{
    sort(L,L+n,cmp);//按极角排序
    head = , tail = ;
    int cnt = ;
    for(int i =  ; i < n- ; i++){
        Vector v1 = L[i].b - L[i].a, v2 = L[i+].b - L[i+].a;//?
        if(fabs(getangle(v1) - getangle(v2)) < eps){
            continue;
        }
        L[cnt++] = L[i];
    }
    L[cnt++] = L[n-];

    /////正片开始
    for(int i =  ; i < cnt ; i++){//注意下标
        while(tail - head >  && on_right(q[tail-], q[tail-], L[i]))    tail--;//符合onright规则，删去上一条直线
        while(tail - head >  && on_right(q[head], q[head+], L[i]))    head++;//头部同理
        q[tail++] = L[i];
    }
    //换个角度看世界，最后检验q[head]以及q[tail-1],q[tail]实际为空
    while(tail - head >  && on_right(q[tail-], q[tail-], q[head]))    tail--;
    while(tail - head >  && on_right(q[head], q[head+], q[tail-]))    head++;

    if((tail - head) >= )    return true;
    return false;
}

ld get_S()//求初始面积
{
    ld ans = ;
    for(int i =  ; i < n ; i++){
        ans += fabs((node[i] - node[]) * (node[(i+)%n] - node[]));
    }
    return ans / ;
}

ld dis(point a, point b)
{
    ld dx = b.x - a.x, dy = b.y - a.y;
    return sqrtl(dx * dx + dy * dy);
}//两点间距离 

ld get_inS()//内部半平面交 面积
{
    ld res = ;
    int tot = ;
    for(int i = head ; i < tail- ; i++){
        p[tot++] = getnode(q[i], q[i+]);
    }
    p[tot++] = getnode(q[tail-], q[head]);

    for(int i =  ; i < tot- ; i++){
        res += fabs((p[i] - p[]) * (p[i+] - p[]));
    }
    res /= ;

    for(int i =  ; i < tot- ; i++){
        res += dis(p[i], p[i+]) * r;
    }// c * r
    res += dis(p[], p[tot - ]) * r;

    return res + r * r * pi;
}

int main(){
    //freopen("mow.in","r",stdin);
    int T;cin >> T;
    while( T-- )
    {
        cin >> n >> r >> cost1 >> cost2;
        for(int i = n- ; i >=  ; i--){
            cin >> node[i].x >> node[i].y;
        }//本题中为顺时针方向，因而逆序存点，以确保直线左侧为内侧 

        if(check()){//逆
            for(int i =  ; i < n- ; i++){//
                l[i] = line(node[i], node[i+]);
            }
            l[n-] = line(node[n-], node[]);
        }else{//顺
            for(int i =  ; i < n- ; i++){
                l[i] = line(node[i+],node[i]);
            }
            l[n-] = line(node[], node[n-]);
        }//0 ~ n-1

        narrow_polygon();

        ld s = get_S();//草坪面积
        ld ans = s * cost1;//手动割草费用

        if(HPI()){
            ld mowable = get_inS();//机器割草面积
            ans = min(ans, mowable * cost2 + (s - mowable) * cost1);
        }
        cout << fixed << setprecision() << ans << "\n";

    }

    return ;
}

//实现直线平移，勾画出圆心轨迹