Beyond the Void
BYVoid
NOI 2009 描邊
本文正體字版由OpenCC轉換

問題簡述

平面上有一些筆畫,筆畫兩端爲半圓,中部爲矩形,筆畫可以互相重疊。任務是求平面上筆畫的總面積。

問題建模

這是一個計算幾何問題,我們可以把一個筆畫分解看爲一個矩形和兩個圓相互重疊(如下圖所示)。

image image

由於題中給定的是筆畫的兩個端點的座標和半徑,所以首先要求出矩形的四個端點座標。如下圖所示,設矩形兩端中點分別爲A,B,矩形A所在邊上一個頂點爲C。只需求出clip_image006的長度相等的平面法向量clip_image008,然後clip_image008[1]除以clip_image010求出單位向量,再乘以R就能求出clip_image012。用A,B座標分別加減clip_image012[1],即可求出矩形四個頂點的座標。

image

解法 豎線掃描

算法描述
用一條豎線去切割平面,會與上面的矩形或圓相交而且有一些相交的區間,把這些區間求並以後並乘以豎線的寬度,即可近似認爲是豎線覆蓋到的圖形面積。用豎線掃描完整個平面,其面積之和就是平面上圖形面積。掃描的豎線越細,結果就越精確。

image

我們要解決豎線與線段相交,以及豎線與圓相交的問題。

如圖所示,掃描線x=x0與線段AB相交,我們要求出交點D的縱座標。分別過B作平行於x=x0的直線,過A作垂直於x=x0的直線,相交於點C,AC交x=x0與點E。顯然有

clip_image018

所以

clip_image020

於是用E的座標加上clip_image022可算出D座標。

image

如圖是掃描線與圓A相交點C,作AB垂直於掃描線與點B,連接AC。|AC|=R,|AB|爲掃描線橫座標與點A橫座標之差絕對值,根據勾股定理即可求出|BC|,進而求出掃描線與圓的兩交點座標。

在程序實現時,我們需要求出每個“筆畫”的左右邊界橫座標,按照左邊界座標排序所有“筆畫”,並用一個雙向鏈表實現隊列維護當前掃描線可以掃描到的“筆畫”。掃描線向右移動時,判斷是否有新的“筆畫”左界小於等於掃描線橫座標,並加入隊列。同時,當發現一個“筆畫”右界小於掃描線橫座標,則把它從隊列中刪除。

實際測試
時間複雜度爲O(MC),M爲筆畫數,C爲掃描線數。在實際的測試中,前9個測試點掃描豎線寬度設定爲1*10^(-6),均得到了10分,程序運行時間如下表。
測試點 1 2 3 4 5 6 7 8 9
時間() 10 10 34 25 50 27 10 17 335

而第10個測試點由於規模過大,運行400秒僅能得到7分。當我把掃描豎線寬度設定爲5*10^(-8),,並運行了11天零4個小時後,終於獲得了10分。

參考程序
```cpp /* * Problem: NOI2009 path * Author: Guo Jiabao * Time: 2009.10.9 13:50 * State: Solved * Memo: 豎線掃描 */ #include #include #include #include #include #include using namespace std; const int MAXN = 2501; const double INF = 1e100;

struct point { double x,y; point (double tx,double ty) : x(tx),y(ty) {} point () : x(0) , y(0) {} };

struct shape { point c[2],v[4]; double left,right; };

struct itvs { double a; bool s; }isl[MAXN];

struct delist { struct item { item *c[2]; shape *e; item(item *last,shape *te) { c[0] = last; c[1] = NULL; e = te; } }*front,*rear; delist() { front = rear = 0; } void insert(shape *e) { if (rear) rear = rear -> c[1] = new item(rear,e); else front = rear = new item(NULL,e); } void remove(item *i) { if (i -> c[0]) i -> c[0] -> c[1] = i -> c[1]; else front = i -> c[1]; if (i -> c[1]) i -> c[1] -> c[0] = i -> c[0]; else rear = i -> c[0]; delete i; } };

int M,N,ic; double R,Total,Left,Right,miny,maxy; point P[MAXN]; shape A[MAXN]; delist DL;

void operator *= (point &a,double k) { a.x *=k; a.y *=k; }

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); }

inline double MIN(double a,double b) { return a<b?a:b; }

inline double MAX(double a,double b) { return a>b?a:b; }

inline double ABS(double a) { return a<0?-a:a; }

inline double dist(point a,point b) { return sqrt((a.x-b.x)(a.x-b.x) + (a.y-b.y)(a.y-b.y)); }

void init() { int i,a,b; freopen(“path.in”,“r”,stdin); freopen(“path.out”,“w”,stdout); scanf("%d”,&N); for (i=1;i<=N;i++) scanf("%lf%lf”,&P[i].x,&P[i].y); scanf("%d”,&M); for (i=1;i<=M;i++) { scanf("%d%d”,&a,&b); A[i].c[0] = P[a]; A[i].c[1] = P[b]; } scanf("%lf”,&R); }

void count_point() { int i; point v,v1,v2; Left = INF; Right = -INF; for (i=1;i<=M;i++) { v = A[i].c[1] - A[i].c[0]; v1.x = v.y; v1.y = -v.x; v1 *= R / dist(A[i].c[0],A[i].c[1]); A[i].v[0] = A[i].c[0] + v1; A[i].v[1] = A[i].c[0] - v1; A[i].v[2] = A[i].c[1] + v1; A[i].v[3] = A[i].c[1] - v1;

    A[i].left = MIN(A[i].c[0].x-R,A[i].c[1].x-R);
    if (A[i].left < Left)
        Left = A[i].left;
    
    A[i].right = MAX(A[i].c[0].x+R,A[i].c[1].x+R);
    if (A[i].right > Right)
        Right = A[i].right;
}

}

inline int cmp(const void *a,const void *b) { return ((shape *)a) -> left < ((shape *)b) -> left ? -1 : 1; }

inline int cmp1(const void *a,const void *b) { if ( ((itvs *)a) -> a < ((itvs *)b) -> a ) return -1; else if ( ((itvs *)a) -> a > ((itvs *)b) -> a ) return 1; if (((itvs *)a) -> s) return -1; return 1; }

void insert_ivl(double s,double t) { if (s >= t) return; isl[++ic].a = s; isl[ic].s = true; isl[++ic].a = t; isl[ic].s = false; }

double ivl_merge() { int i,lv=0; double s,t=0; qsort(isl+1,ic,sizeof(isl[0]),cmp1); for (i=1;i<=ic;i++) { if (isl[i].s) { if (++lv == 1) s = isl[i].a; } else { if (–lv == 0) t += isl[i].a - s; } } return t; }

inline bool isin(double x,double x1,double x2) { if (x1 < x2) return x1 <= x && x <= x2; else return x2 <= x && x <= x1; }

void cross_segment(double x,point a,point b) { if (b.x == a.x || !isin(x,a.x,b.x)) return; double y = (b.y - a.y) * (x - a.x) / (b.x - a.x); y += a.y; if (y < miny) miny = y; if (y > maxy) maxy = y; }

void scan() { const double dx = 1e-6; double x; double per,lper=0; int p=0; for (x=Left;x<=Right;x+=dx) { ////// per = (x - Left)/(Right - Left) * 100; if (per - lper >=1) { lper = per; fprintf(stderr,"%.0lf percent\n”,per); } ////// while (p+1 <=M && A[p+1].left <= x) DL.insert(&A[++p]); ic = 0; for (delist :: item *it = DL.front;it;it = it ->c[1]) { shape *S; for (;it;) { S = it -> e; if (x <= S->right) break; delist :: item *itn = it -> c[1]; DL.remove(it); it = itn; } if (!it) break;

        //Cross circle
        for (int i=0;i<2;i++)
        {
            if (S->c[i].x-R < x && x < S->c[i].x+R)
            {
                double tx,ty;
                tx = ABS(x - S->c[i].x);
                ty = sqrt(R*R - tx*tx);
                insert_ivl(S->c[i].y - ty , S->c[i].y + ty);
            }
        }
        
        miny = INF; maxy = -INF;
        cross_segment(x,S->v[0],S->v[1]);
        cross_segment(x,S->v[2],S->v[3]);
        cross_segment(x,S->v[0],S->v[2]);
        cross_segment(x,S->v[1],S->v[3]);
        insert_ivl(miny,maxy);
    }
    double t = ivl_merge();
    Total += t * dx;
}

}

void solve() { count_point(); qsort(A+1,M,sizeof(A[0]),cmp); scan(); printf("%.12lf\n”,Total); }

int main() { int c_start,c_end; c_start = clock(); init(); solve(); c_end = clock(); cerr «"Running Time : " « double(c_end - c_start) / CLOCKS_PER_SEC « endl; return 0; }


上次修改時間 2017-05-26

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