World Final 1999 poj 1873 The Fortified Forest 狀壓列舉 凸包
阿新 • • 發佈：2019-01-23
2^15 的複雜度 直接列舉集合
但是...
poj tle uvaliva上ac的程式碼：
//#include<iostream> #include<cmath> #include<cstdio> #include<algorithm> #include<vector> #include<cstring> const double eps=1e-10; const double PI=acos(-1.0); const double INF=0x3fffffff; using namespace std; struct Point{ double x; double y; Point(double x=0,double y=0):x(x),y(y){} }; int dcmp(double x) {return (x>eps)-(x<-eps); } int sgn(double x) {return (x>eps)-(x<-eps); } typedef Point Vector; 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);} //ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;} bool operator< (const Point &A,const Point &B) { return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0); } 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 Cross(Vector A,Vector B) {return A.x*B.y-B.x*A.y; } double Length(Vector A) { return sqrt(Dot(A, A));} double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));} double Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);} Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));} Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);} Point GetLineIntersection(Point P,Vector v,Point Q,Vector w) { Vector u=P-Q; double t=Cross(w, u)/Cross(v,w); return P+v*t; } double DistanceToLine(Point P,Point A,Point B) { Vector v1=P-A; Vector v2=B-A; return fabs(Cross(v1,v2))/Length(v2); } double DistanceToSegment(Point P,Point A,Point B) { if(A==B) return Length(P-A); Vector v1=B-A; Vector v2=P-A; Vector v3=P-B; if(dcmp(Dot(v1,v2))==-1) return Length(v2); else if(Dot(v1,v3)>0) return Length(v3); else return DistanceToLine(P, A, B); } Point GetLineProjection(Point P,Point A,Point B) { Vector v=B-A; Vector v1=P-A; double t=Dot(v,v1)/Dot(v,v); return A+v*t; } bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2) { double c1=Cross(b1-a1, a2-a1); double c2=Cross(b2-a1, a2-a1); double c3=Cross(a1-b1, b2-b1); double c4=Cross(a2-b1, b2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ; } bool OnSegment(Point P,Point A,Point B) { return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0; } double PolygonArea(Point *p,int n) { double area=0; for(int i=1;i<n-1;i++) { area+=Cross(p[i]-p[0], p[i+1]-p[0]); } return area/2; } Point read_point() { Point P; scanf("%lf%lf",&P.x,&P.y); return P; } // ---------------與圓有關的-------- struct Circle { Point c; double r; Circle(Point c=Point(0,0),double r=0):c(c),r(r) {} Point point(double a) { return Point(c.x+r*cos(a),c.y+r*sin(a)); } }; struct Line { Point p; Vector v; Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {} Point point(double t) { return Point(p+v*t); } }; int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol) { double a=L.v.x; double b=L.p.x-C.c.x; double c=L.v.y; double d=L.p.y-C.c.y; double e=a*a+c*c; double f=2*(a*b+c*d); double g=b*b+d*d-C.r*C.r; double delta=f*f-4*e*g; if(dcmp(delta)<0) return 0; if(dcmp(delta)==0) { t1=t2=-f/(2*e); sol.push_back(L.point(t1)); return 1; } else { t1=(-f-sqrt(delta))/(2*e); t2=(-f+sqrt(delta))/(2*e); sol.push_back(L.point(t1)); sol.push_back(L.point(t2)); return 2; } } // 向量極角公式 double angle(Vector v) {return atan2(v.y,v.x);} int getCircleCircleIntersection(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; // 重合 else return 0; // 內含 0 個公共點 } if(dcmp(C1.r+C2.r-d)<0) return 0; // 外離 if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0; // 內含 double a=angle(C2.c-C1.c); double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); Point p1=C1.point(a-da); Point p2=C1.point(a+da); sol.push_back(p1); if(p1==p2) return 1; // 相切 else { sol.push_back(p2); return 2; } } // 求點到圓的切線 int getTangents(Point p,Circle C,Vector *v) { Vector u=C.c-p; double dist=Length(u); if(dcmp(dist-C.r)<0) return 0; else if(dcmp(dist-C.r)==0) { v[0]=Rotate(u,PI/2); return 1; } else { double ang=asin(C.r/dist); v[0]=Rotate(u,-ang); v[1]=Rotate(u,+ang); return 2; } } // 求兩圓公切線 int getTangents(Circle A,Circle B,Point *a,Point *b) { int cnt=0; if(A.r<B.r) { swap(A,B); swap(a, b); // 有時需標記 } double d=Length(A.c-B.c); double rdiff=A.r-B.r; double rsum=A.r+B.r; if(dcmp(d-rdiff)<0) return 0; // 內含 double base=angle(B.c-A.c); if(dcmp(d)==0&&dcmp(rdiff)==0) return -1 ; // 重合 無窮多條切線 if(dcmp(d-rdiff)==0) // 內切 外公切線 { a[cnt]=A.point(base); b[cnt]=B.point(base); cnt++; return 1; } // 有外公切線的情形 double ang=acos(rdiff/d); a[cnt]=A.point(base+ang); b[cnt]=B.point(base+ang); cnt++; a[cnt]=A.point(base-ang); b[cnt]=B.point(base-ang); cnt++; if(dcmp(d-rsum)==0) // 外切 有內公切線 { a[cnt]=A.point(base); b[cnt]=B.point(base+PI); cnt++; } else if(dcmp(d-rsum)>0) // 外離 又有兩條外公切線 { double ang_in=acos(rsum/d); a[cnt]=A.point(base+ang_in); b[cnt]=B.point(base+ang_in+PI); cnt++; a[cnt]=A.point(base-ang_in); b[cnt]=B.point(base-ang_in+PI); cnt++; } return cnt; } // 幾何演算法模板 int isPointInPolygon(Point p,Point * poly,int n) { int wn=0; for(int i=0;i<n;i++) { if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1; int k=dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i])); int d1=dcmp(poly[i].y-p.y); int d2=dcmp(poly[(i+1)%n].y-p.y); if(k>0&&d1<=0&&d2>0) wn++; if(k<0&&d2<=0&&d1>0) wn--; } if(wn!=0) return 1; else return 0; } // Andrew 演算法求凸包 int ConvexHull(Point *p,int n,Point *ch) { int m=0; sort(p,p+n); n=unique(p, p+n)-p; for(int i=0;i<n;i++) { while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--; ch[m++]=p[i]; } int k=m; for(int i=n-2;i>=0;i--) { while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--; ch[m++]=p[i]; } if(n>1) m--; return m; } Point p[100]; int value[100]; int l[100]; Point ch[100]; Point fence[100]; Point remain[100]; vector<int> ID[(1<<15)+10]; int ok[1<<15]; double cost[1<<15]; double extra[1<<15]; int n; void init() { for(int i=0;i<(1<<n);i++) { ID[i].clear(); } memset(ok, 0, sizeof(ok)); } int main() { bool first=1; int index=0; while(scanf("%d",&n)==1&&n) { init(); for(int i=0;i<n;i++) { p[i]=read_point(); scanf("%d%d",&value[i],&l[i]); } int all=(1<<n)-1; double minCost=INF; int id=0; for(int i=0;i<=all;i++) { int fence_cnt=0; int remain_cnt=0; int value_sum=0; int l_sum=0; for(int j=0;j<n;j++) { if((1<<j)&i) { fence[fence_cnt++]=p[j]; value_sum+=value[j]; l_sum+=l[j]; ID[i].push_back(j); } else { remain[remain_cnt++]=p[j]; } } // 剪枝 if(value_sum>minCost) continue; double perimeter=0; if(remain_cnt==1) perimeter=0; else if(remain_cnt==2) { perimeter=2*Length(remain[1]-remain[0]); } else { int m=ConvexHull(remain, remain_cnt, ch); for(int i=0;i<m;i++) { perimeter+=Length(ch[(i+1)%m]-ch[i]); } } if(perimeter<=l_sum) { ok[i]=1; cost[i]=value_sum; extra[i]=l_sum-perimeter; } if(ok[i]) { if(value_sum<minCost) { id=i; minCost=value_sum; } else if(dcmp(value_sum-minCost)==0) { if(ID[i].size()<ID[id].size()) { id=i; } } } } if(first) { first=0; } else{ puts(""); } printf("Forest %d\n",++index); printf("Cut these trees:"); for(int i=0;i<ID[id].size();i++) { printf(" %d",ID[id][i]+1); } puts(""); printf("Extra wood: "); printf("%.2f\n",extra[id]); } }
最後發現原因是每個幾何都存一遍vector裡面 效率.. 被卡常數了
優化1 當value_sum>當前的value_sum 就不去求凸包了 一定不會更新答案（剪枝）
2 根本不用儲存 記住集合的id 就行 最後直接列印這個集合
3 如果列舉集合的時候倒著列舉 能更快一點程式碼：（poj ac）
#include<iostream> #include<cmath> #include<cstdio> #include<algorithm> #include<vector> #include<cstring> const double eps=1e-6; const double PI=acos(-1.0); const double INF=0x3fffffff; using namespace std; struct Point{ double x; double y; Point(double x=0,double y=0):x(x),y(y){} }; int dcmp(double x) {return (x>eps)-(x<-eps); } int sgn(double x) {return (x>eps)-(x<-eps); } typedef Point Vector; 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);} //ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;} bool operator< (const Point &A,const Point &B) { return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0); } 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 Cross(Vector A,Vector B) {return A.x*B.y-B.x*A.y; } double Length(Vector A) { return sqrt(Dot(A, A));} double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));} double Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);} Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));} Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);} Point GetLineIntersection(Point P,Vector v,Point Q,Vector w) { Vector u=P-Q; double t=Cross(w, u)/Cross(v,w); return P+v*t; } double DistanceToLine(Point P,Point A,Point B) { Vector v1=P-A; Vector v2=B-A; return fabs(Cross(v1,v2))/Length(v2); } double DistanceToSegment(Point P,Point A,Point B) { if(A==B) return Length(P-A); Vector v1=B-A; Vector v2=P-A; Vector v3=P-B; if(dcmp(Dot(v1,v2))==-1) return Length(v2); else if(Dot(v1,v3)>0) return Length(v3); else return DistanceToLine(P, A, B); } Point GetLineProjection(Point P,Point A,Point B) { Vector v=B-A; Vector v1=P-A; double t=Dot(v,v1)/Dot(v,v); return A+v*t; } bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2) { double c1=Cross(b1-a1, a2-a1); double c2=Cross(b2-a1, a2-a1); double c3=Cross(a1-b1, b2-b1); double c4=Cross(a2-b1, b2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ; } bool OnSegment(Point P,Point A,Point B) { return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0; } double PolygonArea(Point *p,int n) { double area=0; for(int i=1;i<n-1;i++) { area+=Cross(p[i]-p[0], p[i+1]-p[0]); } return area/2; } Point read_point() { Point P; scanf("%lf%lf",&P.x,&P.y); return P; } // ---------------與圓有關的-------- struct Circle { Point c; double r; Circle(Point c=Point(0,0),double r=0):c(c),r(r) {} Point point(double a) { return Point(c.x+r*cos(a),c.y+r*sin(a)); } }; struct Line { Point p; Vector v; Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {} Point point(double t) { return Point(p+v*t); } }; int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol) { double a=L.v.x; double b=L.p.x-C.c.x; double c=L.v.y; double d=L.p.y-C.c.y; double e=a*a+c*c; double f=2*(a*b+c*d); double g=b*b+d*d-C.r*C.r; double delta=f*f-4*e*g; if(dcmp(delta)<0) return 0; if(dcmp(delta)==0) { t1=t2=-f/(2*e); sol.push_back(L.point(t1)); return 1; } else { t1=(-f-sqrt(delta))/(2*e); t2=(-f+sqrt(delta))/(2*e); sol.push_back(L.point(t1)); sol.push_back(L.point(t2)); return 2; } } // 向量極角公式 double angle(Vector v) {return atan2(v.y,v.x);} int getCircleCircleIntersection(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; // 重合 else return 0; // 內含 0 個公共點 } if(dcmp(C1.r+C2.r-d)<0) return 0; // 外離 if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0; // 內含 double a=angle(C2.c-C1.c); double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); Point p1=C1.point(a-da); Point p2=C1.point(a+da); sol.push_back(p1); if(p1==p2) return 1; // 相切 else { sol.push_back(p2); return 2; } } // 求點到圓的切線 int getTangents(Point p,Circle C,Vector *v) { Vector u=C.c-p; double dist=Length(u); if(dcmp(dist-C.r)<0) return 0; else if(dcmp(dist-C.r)==0) { v[0]=Rotate(u,PI/2); return 1; } else { double ang=asin(C.r/dist); v[0]=Rotate(u,-ang); v[1]=Rotate(u,+ang); return 2; } } // 求兩圓公切線 int getTangents(Circle A,Circle B,Point *a,Point *b) { int cnt=0; if(A.r<B.r) { swap(A,B); swap(a, b); // 有時需標記 } double d=Length(A.c-B.c); double rdiff=A.r-B.r; double rsum=A.r+B.r; if(dcmp(d-rdiff)<0) return 0; // 內含 double base=angle(B.c-A.c); if(dcmp(d)==0&&dcmp(rdiff)==0) return -1 ; // 重合 無窮多條切線 if(dcmp(d-rdiff)==0) // 內切 外公切線 { a[cnt]=A.point(base); b[cnt]=B.point(base); cnt++; return 1; } // 有外公切線的情形 double ang=acos(rdiff/d); a[cnt]=A.point(base+ang); b[cnt]=B.point(base+ang); cnt++; a[cnt]=A.point(base-ang); b[cnt]=B.point(base-ang); cnt++; if(dcmp(d-rsum)==0) // 外切 有內公切線 { a[cnt]=A.point(base); b[cnt]=B.point(base+PI); cnt++; } else if(dcmp(d-rsum)>0) // 外離 又有兩條外公切線 { double ang_in=acos(rsum/d); a[cnt]=A.point(base+ang_in); b[cnt]=B.point(base+ang_in+PI); cnt++; a[cnt]=A.point(base-ang_in); b[cnt]=B.point(base-ang_in+PI); cnt++; } return cnt; } // 幾何演算法模板 int isPointInPolygon(Point p,Point * poly,int n) { int wn=0; for(int i=0;i<n;i++) { if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1; int k=dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i])); int d1=dcmp(poly[i].y-p.y); int d2=dcmp(poly[(i+1)%n].y-p.y); if(k>0&&d1<=0&&d2>0) wn++; if(k<0&&d2<=0&&d1>0) wn--; } if(wn!=0) return 1; else return 0; } // Andrew 演算法求凸包 int ConvexHull(Point *p,int n,Point *ch) { int m=0; sort(p,p+n); n=unique(p, p+n)-p; for(int i=0;i<n;i++) { while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--; ch[m++]=p[i]; } int k=m; for(int i=n-2;i>=0;i--) { while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--; ch[m++]=p[i]; } if(n>1) m--; return m; } Point p[100]; int value[100]; int l[100]; Point ch[100]; Point fence[100]; Point remain[100]; const int maxn=1<<15; double extra[maxn]; int n; //vector<int> ID[(1<<15)+10]; int main() { bool first=1; int index=0; while(scanf("%d",&n)==1&&n) { for(int i=0;i<n;i++) { p[i]=read_point(); scanf("%d%d",&value[i],&l[i]); } int all=(1<<n)-1; int minCost=INF; int id=0; int minTree=INF; for(int i=all;i>=0;i--) { int fence_cnt=0; int remain_cnt=0; int value_sum=0; double l_sum=0; for(int j=0;j<n;j++) { if((1<<j)&i) { fence[fence_cnt++]=p[j]; value_sum+=value[j]; l_sum+=l[j]; // ID[i].push_back(j); } else { remain[remain_cnt++]=p[j]; } } // 剪枝 if(value_sum>minCost) continue; double perimeter=0; if(remain_cnt==1) perimeter=0; else if(remain_cnt==2) { perimeter=2*Length(remain[1]-remain[0]); } else { int m=ConvexHull(remain, remain_cnt, ch); for(int i=0;i<m;i++) { perimeter+=Length(ch[(i+1)%m]-ch[i]); } } bool ok=0; if(dcmp(perimeter-l_sum)<=0) { ok=1; extra[i]=l_sum-perimeter; } if(ok) { if(value_sum<minCost) { id=i; minCost=value_sum; int temp=0; for(int j=0;j<n;j++) { if((1<<j)&i) temp++; } minTree=temp; } else if(value_sum==minCost) { int temp=0; for(int j=0;j<n;j++) { if((1<<j)&i) temp++; } if(temp<=minTree) { id=i; } } } } if(first) { first=0; } else { puts(""); } printf("Forest %d\n",++index); printf("Cut these trees:"); for(int i=0;i<n;i++) { if((1<<i)&id) printf(" %d",i+1); } // for(int i=0;i<ID[id].size();i++) // { // printf(" %d",ID[id][i]+1); // } puts(""); printf("Extra wood: "); printf("%.2f\n",extra[id]); } return 0; }