圆的k次面积并
搬运别人的
https://vjudge.net/problem/SPOJ-CIRUT
//china no.1 #pragma comment(linker, "/STACK:1024000000,1024000000") #include <vector> #include <iostream> #include <string> #include <map> #include <stack> #include <cstring> #include <queue> #include <list> #include <stdio.h> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <iomanip> #include <cctype> #include <sstream> #include <functional> #include <stdlib.h> #include <time.h> #include <bitset> using namespace std; #define pi acos(-1) #define PI acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); #define db double typedef long long LL; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=1e3+10; const int maxx=1e6+100; const double EPS=1e-8; const double eps=1e-8; const int mod=10000007; template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} template <class T> inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;} while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;} inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false; while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;} else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}} if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}} if(IsN) num=-num;return true;} void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');} void print(LL a){ Out(a),puts("");} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); //cerr << "run time is " << clock() << endl; /*struct Point { double x, y; Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数 Point(double x = 0, double y = 0) : x(x), y(y) { } inline void input() { scanf("%lf%lf",&x,&y); } inline void print() { printf("%.6lf %.6lf\n",x,y); } };*/ db sqr(db x) { return x*x; } int dcmp(double x) { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; } struct Circle { double x, y, r, angle; int d; Circle(){} Circle(double xx, double yy, double ang = 0, int t = 0) { x = xx; y = yy; angle = ang; d = t; } void get() { scanf("%lf%lf%lf", &x, &y, &r); d = 1; } }; Circle cir[maxn],tp[maxn*2]; double area[maxn]; double dis(Circle a,Circle b) { return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y)); } double cross(Circle p0,Circle p1,Circle p2) { return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x); } //圆相交 int CirCrossCir(Circle p1, double r1,Circle p2, double r2,Circle &cp1,Circle &cp2) { double mx = p2.x - p1.x, sx = p2.x + p1.x, mx2 = mx * mx; double my = p2.y - p1.y, sy = p2.y + p1.y, my2 = my * my; double sq = mx2 + my2, d = -(sq - sqr(r1 - r2)) * (sq - sqr(r1 + r2)); if (d + eps < 0) return 0; if (d < eps) d = 0; else d = sqrt(d); double x = mx * ((r1 + r2) * (r1 - r2) + mx * sx) + sx * my2; double y = my * ((r1 + r2) * (r1 - r2) + my * sy) + sy * mx2; double dx = mx * d, dy = my * d; sq *= 2; cp1.x = (x - dy) / sq; cp1.y = (y + dx) / sq; cp2.x = (x + dy) / sq; cp2.y = (y - dx) / sq; if (d > eps) return 2; else return 1; } bool circmp(const Circle& u, const Circle& v) { return dcmp(u.r - v.r) < 0; } bool cmp(const Circle& u, const Circle& v) { if (dcmp(u.angle - v.angle)) return u.angle < v.angle; return u.d > v.d; } //0.5*r*r*(K-sin(K)) double calc(Circle cir,Circle cp1,Circle cp2) { double ans = (cp2.angle - cp1.angle) * sqr(cir.r) - cross(cir, cp1, cp2) + cross(Circle(0, 0), cp1, cp2); return ans / 2; } void CirUnion(Circle cir[], int n) { Circle cp1, cp2; sort(cir, cir + n, circmp); for (int i = 0; i < n; ++i) for (int j = i + 1; j < n; ++j) if (dcmp(dis(cir[i], cir[j]) + cir[i].r - cir[j].r) <= 0) cir[i].d++; for (int i = 0; i < n; ++i) { int tn = 0, cnt = 0; for (int j = 0; j < n; ++j) { if (i == j) continue; if (CirCrossCir(cir[i], cir[i].r, cir[j], cir[j].r, cp2, cp1) < 2) continue; cp1.angle = atan2(cp1.y - cir[i].y, cp1.x - cir[i].x); cp2.angle = atan2(cp2.y - cir[i].y, cp2.x - cir[i].x); cp1.d = 1; tp[tn++] = cp1; cp2.d = -1; tp[tn++] = cp2; if (dcmp(cp1.angle - cp2.angle) > 0) cnt++; } tp[tn++] = Circle(cir[i].x - cir[i].r, cir[i].y, pi, -cnt); tp[tn++] = Circle(cir[i].x - cir[i].r, cir[i].y, -pi, cnt); sort(tp, tp + tn, cmp); int p, s = cir[i].d + tp[0].d; for (int j = 1; j < tn; ++j) { p = s; s += tp[j].d; area[p] += calc(cir[i], tp[j - 1], tp[j]); } } } int n; void solve() { for(int i=0;i<n;i++) cir[i].get(); me(area,0); CirUnion(cir,n); for(int i=1;i<=n;i++) { area[i]-=area[i+1]; printf("[%d] = %.3f\n", i, area[i]); } } int main() { while(scanf("%d",&n)!=EOF) solve(); }
作者:weeping
出处:www.cnblogs.com/weeping/
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