不说了。。。说多了都是泪。。。从昨天下午一直wa到现在,直到刚刚才让人帮我找到所谓的“bug”,其实也算不上bug。。。
这个题的思路就是:找出平面上的所有点:所有圆的交点以及所有圆的圆心。然后依次判断两点是否连通,连通的话两点距离便是其欧几里得距离。这样建完图之后直接跑s->t最短路就行了。。
两点连通?也就是说这两点连成的线段,一直在圆内,任意圆都行。如何判断呢,求出该线段与所有圆的所有交点,排序后将其分段,依次判断每一段是否在任意圆内。这个么,在分段后,判断每一段的中点是否在圆内就行了。
这题并不是很难啊?我的bug在哪??刚开始我是将所有圆的圆心都加到平面点集中,wa到死有木有。。。刚刚改了改,只加第一个跟最后一个圆的圆心。。立A。。为什么?
#include<algorithm> #include<iostream> #include<cstring> #include<fstream> #include<sstream> #include<vector> #include<string> #include<cstdio> #include<bitset> #include<queue> #include<stack> #include<cmath> #include<map> #include<set> #define FF(i, a, b) for(int i=a; i<b; i++) #define FD(i, a, b) for(int i=a; i>=b; i--) #define REP(i, n) for(int i=0; i<n; i++) #define CLR(a, b) memset(a, b, sizeof(a)) #define LL long long #define PB push_back #define eps 1e-10 #define debug puts("**debug**") using namespace std; const double PI = acos(-1); const double INF = 1e20; const int maxn = 11111; struct Point { double x, y; Point (double x=0, double y=0):x(x), y(y) {} }; typedef Point Vector; struct Circle { Point c; double r; Circle() {} Circle(Point c, double r) : c(c), r(r) {} Point point(double a) { return Point(c.x+cos(a)*r, c.y+sin(a)*r); } }; struct Line { Point p; Vector v; double ang; Line(){} Line(Point p, Vector v) : p(p), v(v) {ang = atan2(v.y, v.x); } Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); } bool operator < (const Line& L) const { return ang < L.ang; } }; template <class T> T sqr(T x) { return x * x;} Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); } Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); } bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; } bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; } int dcmp(double x) { if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1; } bool operator == (const Point& a, const Point& b){ return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;} double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); } double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } Vector vecunit(Vector x){ return x / Length(x);} Vector normal(Vector x) { return Point(-x.y, x.x) / Length(x);} double angel(Vector v) { return atan2(v.y, v.x); } bool OnSegment(Point p, Point a1, Point a2) { return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; } bool InCircle(Point x, Circle c) { return dcmp(sqr(c.r) - sqr(Length(c.c - x))) > 0;} double DistanceToSeg(Point p, Point a, Point b) { if(a == b) return Length(p-a); Vector v1 = b-a, v2 = p-a, v3 = p-b; if(dcmp(Dot(v1, v2)) < 0) return Length(v2); if(dcmp(Dot(v1, v3)) > 0) return Length(v3); return fabs(Cross(v1, v2)) / Length(v1); } void getLineABC(Point A, Point B, double& a, double& b, double& c) { a = A.y-B.y, b = B.x-A.x, c = A.x*B.y-A.y*B.x; } Point GetIntersection(Line a, Line b) { Vector u = a.p-b.p; double t = Cross(b.v, u) / Cross(a.v, b.v); return a.p + a.v*t; } int getCCintersection(Circle C1, Circle C2, vector<Point>& sol) { double d = Length(C1.c - C2.c); if(dcmp(d) == 0) { if(dcmp(C1.r - C2.r) == 0) return -1; return 0; } if(dcmp(C1.r + C2.r - d) < 0) return 0; if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0; double a = angel(C2.c - C1.c); double da = acos((sqr(C1.r) + sqr(d) - sqr(C2.r)) / (2*C1.r*d)); Point p1 = C1.point(a-da), p2 = C1.point(a+da); sol.PB(p1); if(p1 == p2) return 1; sol.PB(p2); return 2; } int getSegCircleIntersection(Line L, Circle C, Point* sol) { Vector nor = normal(L.v); Line pl = Line(C.c, nor); Point ip = GetIntersection(pl, L); double dis = Length(ip - C.c); if (dcmp(dis - C.r) > 0) return 0; Point dxy = vecunit(L.v) * sqrt(sqr(C.r) - sqr(dis)); int ret = 0; sol[ret] = ip + dxy; if (OnSegment(sol[ret], L.p, L.point(1))) ret++; sol[ret] = ip - dxy; if (OnSegment(sol[ret], L.p, L.point(1))) ret++; return ret; } int T, n; vector<Point> sol; Circle C[30]; Point s, t; bool vis[maxn]; bool check(Point A, Point B) { vector<Point> gank; gank.PB(A); gank.PB(B); Point roshan[2]; REP(i, n) { //线段ab与所有圆的交点 int m = getSegCircleIntersection(Line(A, B-A), C[i], roshan); if(m == 1) gank.PB(roshan[0]); if(m == 2) gank.PB(roshan[0]), gank.PB(roshan[1]); } sort(gank.begin(), gank.end()); int nc = gank.size(); REP(i, nc-1) { Point mid = (gank[i] + gank[i+1]) / 2.0; //重点跳过 if(mid == gank[i]) continue; bool flag = 0; REP(j, n) if(InCircle(mid, C[j]))//分段中点是否被任意圆覆盖 { flag = 1; break; } if(!flag) return false; } return true; } struct Heap { double d; int u; bool operator < (const Heap& rhs) const { return dcmp(d-rhs.d) > 0; } }; struct Edge { int from, to; double dist; }; vector<Edge> edges; vector<int> G[maxn]; bool done[maxn]; double d[maxn]; double dij(int s, int t, int n) { priority_queue<Heap> q; q.push((Heap){0.0, s}); REP(i, n) d[i] = INF; d[s] = 0.0; CLR(done, 0); while(!q.empty()) { Heap x = q.top(); q.pop(); int u = x.u, nc = G[u].size(); if(done[u]) continue; done[u] = 1; REP(i, nc) { Edge& e = edges[G[u][i]]; if(dcmp(d[e.to]-d[u]-e.dist) > 0) { d[e.to] = d[u] + e.dist; q.push((Heap){d[e.to], e.to}); } } } return d[t]; } void add(int from, int to, double dist) { edges.PB((Edge){from, to, dist}); edges.PB((Edge){to, from, dist}); int nc = edges.size(); G[from].PB(nc-2); G[to].PB(nc-1); } void init() { REP(i, maxn) G[i].clear(); sol.clear(); edges.clear(); } int main() { scanf("%d", &T); FF(kase, 1, T+1) { init(); scanf("%d", &n); //sol存平面上所有点 REP(i, n) { scanf("%lf%lf%lf", &C[i].c.x, &C[i].c.y, &C[i].r); if(i == 0 || i == n-1) sol.PB(C[i].c);//加入圆心.... } REP(i, n) FF(j, i+1, n) getCCintersection(C[i], C[j], sol);//所有圆的交点 //去重 sort(sol.begin(), sol.end()); int nc = unique(sol.begin(), sol.end()) - sol.begin(); int s, t; REP(i, nc) { FF(j, i+1, nc) if(check(sol[i], sol[j])) //判断两点是否连通 add(i, j, Length(sol[i] - sol[j])); //最短路的起始点 if(sol[i] == C[0].c) s = i; if(sol[i] == C[n-1].c) t = i; } double ans = dij(s, t, nc); printf("Case %d: ", kase); if(dcmp(ans-INF) >= 0) puts("No such path."); else printf("%.4lf\n", ans); } return 0; }