hdu6755 Mow
- 2020 年 7 月 25 日
- 筆記
- 2020杭電暑期集訓
半平面交 模擬 雙端隊列
慢慢啃
人生第一次程式碼過兩百行啊…加油加油
未來のために 頑張ってください
#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(-1);
const ld eps = 1e-10;//控制精度
int n;
int head, tail;
ld r, cost1, cost2;
typedef struct Grid
{//棋盤
ld x,y;
Grid(double a = 0, double b = 0){x = a, y = b;}//構造函數,結構體對象創建時執行
} point , Vector ;//point:類型為Grid的數據類型;Vector與point結構相同,數據類型名稱不同
typedef struct Net
{
point a,b;
Net() {}//重載構造函數,目的是以形如line x;的方式定義直線,避免定義時確定參數的複雜性
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需大寫,避免與vector混淆
{
return Vector(b.x - a.x, b.y - a.y);
}//矢量定義
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);//返回一個(-pi,pi]的角度
}//獲得某個矢量的極角
ld getangle(line t)
{
return atan2(t.b.y - t.a.y, t.b.x - t.a.x);
}//獲得某條直線的極角
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) >= 0);//靠左邊的丟在後面,方便雙端隊列去重
}
return A < B;
}//決定根據極角排序的直線的優先順序
bool check()
{
ld ans = 0;
for(int i = 1 ; i < n-1 ; i++){
ans += (node[i] - node[0]) * (node[i+1] - node[0]);
}
if(ans < 0) return false;
return true;
}//檢查point輸入順序進行調整 true為逆時針
point getnode(line a, line b)//求兩直線(線段)交點,一般式三參數暴力求解
{
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移至右側變號
return point((c2 * b1 - b2 * c1) / D, (c1 * a2 - a1 * c2) / D);
}
bool on_right(line a, line b, line c)//觀察a、b交點與c的關係
{
point n = getnode(a, b);
if((c.b - c.a) * (n - c.a) < 0){//交點在右側/下方
return true;
}
return false;
}
line narrow_line(line x)//實現直線平移,勾畫出圓心軌跡
{
ld dx = x.b.x - x.a.x, dy = x.b.y - x.a.y;
Vector turnv;//旋轉矢量
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;
//求對應坐標
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()//模擬圓心軌跡
{
for(int i = 0 ; i <= n-1 ; i++){
L[i] = narrow_line(l[i]);
}
}
bool HPI()
{
sort(L,L+n,cmp);//按極角排序
head = 0, tail = 0;
int cnt = 0;
for(int i = 0 ; i < n-1 ; i++){
Vector v1 = L[i].b - L[i].a, v2 = L[i+1].b - L[i+1].a;//?
if(fabs(getangle(v1) - getangle(v2)) < eps){
continue;
}
L[cnt++] = L[i];
}
L[cnt++] = L[n-1];
/////正片開始
for(int i = 0 ; i < cnt ; i++){//注意下標
while(tail - head > 1 && on_right(q[tail-1], q[tail-2], L[i])) tail--;//符合onright規則,刪去上一條直線
while(tail - head > 1 && on_right(q[head], q[head+1], L[i])) head++;//頭部同理
q[tail++] = L[i];
}
//換個角度看世界,最後檢驗q[head]以及q[tail-1],q[tail]實際為空
while(tail - head > 1 && on_right(q[tail-1], q[tail-2], q[head])) tail--;
while(tail - head > 1 && on_right(q[head], q[head+1], q[tail-1])) head++;
if((tail - head) >= 3) return true;
return false;
}
ld get_S()//求初始面積
{
ld ans = 0;
for(int i = 1 ; i < n ; i++){
ans += fabs((node[i] - node[0]) * (node[(i+1)%n] - node[0]));
}
return ans / 2;
}
ld dis(point a, point b)
{
ld dx = b.x - a.x, dy = b.y - a.y;
return sqrtl(dx * dx + dy * dy);
}//兩點間距離
ld get_inS()//內部半平面交 面積
{
ld res = 0;
int tot = 0;
for(int i = head ; i < tail-1 ; i++){
p[tot++] = getnode(q[i], q[i+1]);
}
p[tot++] = getnode(q[tail-1], q[head]);
for(int i = 1 ; i < tot-1 ; i++){
res += fabs((p[i] - p[0]) * (p[i+1] - p[0]));
}
res /= 2;
for(int i = 0 ; i < tot-1 ; i++){
res += dis(p[i], p[i+1]) * r;
}// c * r
res += dis(p[0], p[tot - 1]) * 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-1 ; i >= 0 ; i--){
cin >> node[i].x >> node[i].y;
}//本題中為順時針方向,因而逆序存點,以確保直線左側為內側
if(check()){//逆
for(int i = 0 ; i < n-1 ; i++){//
l[i] = line(node[i], node[i+1]);
}
l[n-1] = line(node[n-1], node[0]);
}else{//順
for(int i = 0 ; i < n-1 ; i++){
l[i] = line(node[i+1],node[i]);
}
l[n-1] = line(node[0], node[n-1]);
}//0 ~ n-1
narrow_polygon();
ld s = get_S();//草坪面積
ld ans = s * cost1;//手動割草費用
if(HPI()){
ld mowable = get_inS();//機器割草面積
ans = min(ans, mowable * cost2 + (s - mowable) * cost1);
}
cout << fixed << setprecision(25) << ans << "\n";
}
return 0;
}
//實現直線平移,勾畫出圓心軌跡