题目大意:
一些点在一张无穷图上面,每个点可以控制一些区域,这个区域满足这个点到达这个区域的时间严格小于其他点。求哪些点能够控制无穷面积的区域。
题目思路:
速度小的控制范围一定有限。
速度最大当且仅当在凸包上才能够控制无穷区域。可以通过,任意两个点中垂线为界,左右各控制一半,判断出凸包内的点仅能控制有限区域。
特判:
速度最大且在同一个点上的点均不能控制无穷区域,但是要加入凸包计算。
速度最大为0不能控制无穷区域。
对于共线凸包(Graham),(代码中有解释)
均不能存在重点!可用map判重。
1、按极角坐标序排
缺点:需要将最后一条边上的点逆序排,才能够将最后一边共线点加入凸包。
2、按水平序排。
缺点:若所有点在一条直线上,会产生将所有点入凸包1~n~2的情况,需要特判,当然本题只是用到这些点,无需判断是否重复出现。
极角序:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <utility> #include <stack> #include <queue> #include <map> #include <deque> #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define INF 0x3f3f3f3f using namespace std; const int MAXN = 1010; const double eps = 1e-8; const double PI = acos(-1.0); int tx,ty,tv,maxv,n,N,cas; bool pd[MAXN]; int sgn(double x) { if(fabs(x) < eps) return 0; if(x < 0) return -1; return 1; } struct Point { double x,y; int re; Point(){} Point(double _x, double _y): x(_x),y(_y) {} Point operator -(const Point &B) const { return Point(x-B.x, y-B.y); } Point operator +(const Point &B) const //向量相加 { return Point(x+B.x, y+B.y); } double operator ^(const Point &B) const //叉积 { return x*B.y - y*B.x; } double operator *(const Point &B) const //点积 { return x*B.x + y*B.y; } bool operator ==(const Point &B) const { return fabs(B.x-x)<eps && fabs(B.y-y)<eps; } bool operator !=(const Point &B) const { return !((*this) == B); } double norm()//向量的模 { return sqrt(x*x+y*y); } void transXY(double B) //绕原点逆时针旋转B弧度 { double tx = x, ty = y; x = tx*cos(B) - ty*sin(B); y = tx*sin(B) + ty*cos(B); } void input() //读入只能用double读入 { scanf("%lf%lf",&x,&y); } }; struct Line { Point s,e; Line(){} Line(Point _s, Point _e) { s=_s; e=_e; } }; double dist(Point a, Point b) { return sqrt((a-b)*(a-b)); } //判断点在线段上 bool OnS(Point A, Line a) { return sgn((a.s-A)^(a.e-A)) == 0 && sgn((A.x-a.s.x)*(A.x-a.e.x)) <= 0 && sgn((A.y-a.s.y)*(A.y-a.e.y)) <= 0; } //求凸包 Graham算法 //点的编号0~n-1 //返回凸包结果Stack[0~top-1]为凸包的编号 //一个点或两个点 则凸包为一或二个点 int Stack[MAXN],top; Point vertex[MAXN]; bool Graham_cmp(Point A, Point B) { double tmp=(A-vertex[0])^(B-vertex[0]); if(sgn(tmp) > 0) return 1; if(sgn(tmp) == 0 && sgn(dist(A,vertex[0])-dist(B,vertex[0])) <= 0) return 1; return 0; } void Graham(int n) { int k=0; for(int i=1; i<n; i++) if((vertex[k].y>vertex[i].y) || (vertex[k].y==vertex[i].y && vertex[k].x>vertex[i].x)) k=i; swap(vertex[0], vertex[k]); sort(vertex+1, vertex+n, Graham_cmp); if(n == 1) { top=1; Stack[0]=0; return; } if(n == 2) { top=2; Stack[0]=0; Stack[1]=1; return; } int tmp; for(tmp=n-1; tmp>1 && sgn((vertex[0]-vertex[tmp])^(vertex[0]-vertex[tmp-1])) == 0; tmp--); reverse(vertex+tmp,vertex+n);//最后一条边倒序 Stack[0]=0; Stack[1]=1; top=2; for(int i=2; i<n; i++) { while(top > 1 && (vertex[i] == vertex[Stack[top-1]] || sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) < 0))//相同点只进栈一次 同一条线上的点也进栈 top--; Stack[top++]=i; } } int main() { // freopen("1002.in","r",stdin); // freopen("1002p.out","w",stdout); while(scanf("%d",&N)!=EOF && N) { memset(pd,0,sizeof(pd)); n=0; maxv=-1; for(int i=0; i<N; i++) { scanf("%d%d%d",&tx,&ty,&tv); if(maxv==tv) { vertex[n].x=tx; vertex[n].y=ty; vertex[n].re=i; n++; } else if(maxv<tv) { maxv=tv; n=0; vertex[n].x=tx; vertex[n].y=ty; vertex[n].re=i; n++; } } Graham(n); for(int i=0; i<top; i++) pd[vertex[Stack[i]].re]=1; for(int i=0; i<n; i++)//去掉相同点 for(int j=i+1; j<n; j++) if(vertex[i]==vertex[j]) { pd[vertex[i].re]=0; pd[vertex[j].re]=0; } printf("Case #%d: ",++cas); for(int i=0; i<N; i++) { if(maxv==0) printf("0"); else printf("%d",pd[i]); } printf(" "); } return 0; } /* 0 0 1 0 1 0 1 1 1 2 1 3 1 1 1 */
水平序:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <utility> #include <stack> #include <queue> #include <map> #include <deque> #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define INF 0x3f3f3f3f using namespace std; const int MAXN = 1010; const double eps = 1e-8; const double PI = acos(-1.0); int tx,ty,tv,maxv,n,N,cas; int pd[MAXN]; int sgn(double x) { if(fabs(x) < eps) return 0; if(x < 0) return -1; return 1; } struct Point { double x,y; int re; Point(){} Point(double _x, double _y): x(_x),y(_y) {} Point operator -(const Point &B) const { return Point(x-B.x, y-B.y); } Point operator +(const Point &B) const //向量相加 { return Point(x+B.x, y+B.y); } double operator ^(const Point &B) const //叉积 { return x*B.y - y*B.x; } double operator *(const Point &B) const //点积 { return x*B.x + y*B.y; } bool operator ==(const Point &B) const { return fabs(B.x-x)<eps && fabs(B.y-y)<eps; } bool operator !=(const Point &B) const { return !((*this) == B); } double norm()//向量的模 { return sqrt(x*x+y*y); } void transXY(double B) //绕原点逆时针旋转B弧度 { double tx = x, ty = y; x = tx*cos(B) - ty*sin(B); y = tx*sin(B) + ty*cos(B); } bool operator<(const Point B) const { return(x<B.x || (x==B.x && y<B.y)); } void input() //读入只能用double读入 { scanf("%lf%lf",&x,&y); } }; struct Line { Point s,e; Line(){} Line(Point _s, Point _e) { s=_s; e=_e; } }; double dist(Point a, Point b) { return sqrt((a-b)*(a-b)); } //判断点在线段上 bool OnS(Point A, Line a) { return sgn((a.s-A)^(a.e-A)) == 0 && sgn((A.x-a.s.x)*(A.x-a.e.x)) <= 0 && sgn((A.y-a.s.y)*(A.y-a.e.y)) <= 0; } //求凸包 Graham算法 //点的编号0~n-1 //返回凸包结果Stack[0~top-1]为凸包的编号 //一个点或两个点 则凸包为一或二个点 int Stack[MAXN],top; Point vertex[MAXN]; bool Graham_cmp(Point A, Point B) { return A.y<B.y || (A.y == B.y && A.x<B.x); } void Graham(int n) { sort(vertex, vertex+n, Graham_cmp); top=0; for(int i=0; i<n; i++) { while(top > 1 && sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) < 0)//改为<即可 top--; Stack[top++]=i; } int tmp=top; for(int i=n-2; i>=0; i--) { while(top > tmp && sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) < 0) top--; Stack[top++]=i; } if(n>1) top--; } int main() { freopen("1002.in","r",stdin); freopen("1002p.out","w",stdout); map<Point,int> mapp; while(scanf("%d",&N)!=EOF && N) { memset(pd,-1,sizeof(pd)); n=0; maxv=-1; for(int i=1; i<=N; i++) { scanf("%d%d%d",&tx,&ty,&tv); if(maxv==tv && mapp[Point(tx,ty)]>0) { pd[mapp[Point(tx,ty)]]=0; pd[i]=0; continue; } if(maxv==tv) { vertex[n].x=tx; vertex[n].y=ty; vertex[n].re=i; mapp[vertex[n]]=i; n++; } if(maxv<tv) { mapp.clear(); maxv=tv; n=0; vertex[n].x=tx; vertex[n].y=ty; vertex[n].re=i; mapp[vertex[n]]=i; n++; } } Graham(n); for(int i=0; i<top; i++) if(pd[vertex[Stack[i]].re]==-1) pd[vertex[Stack[i]].re]=1; printf("Case #%d: ",++cas); for(int i=1; i<=N; i++) if(maxv==0) printf("0"); else { if(pd[i]<=0) printf("0"); else printf("1"); } printf(" "); } return 0; } /* 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 1 0 1 1 1 2 1 3 1 1 1 */
水平序优化:(可以解决重点+共线凸包问题)
vis判水平序的点是否访问过,防止一条线的情况。
pd判是否重点,在水平排序后相邻的一定相邻!写的还算漂亮。毕竟map太暴力了。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cstdlib> #include <cmath> #include <utility> #include <vector> #include <queue> #include <map> #include <set> #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)>(y)?(y):(x)) #define MAXN 505 #define eps 1e-4 using namespace std; struct Point{ double x,y; int res; Point(){} Point(double _x, double _y): x(_x),y(_y){} double operator^(Point A) { return x*A.y-A.x*y; } Point operator -(const Point A) const { return Point(x-A.x,y-A.y); } }vertex[MAXN]; int Stack[MAXN],top; int N,n,x,y,v,Case; bool vis[MAXN],pd[MAXN]; inline double dist(Point A) { return sqrt(A.x*A.x+A.y*A.y); } int sgn(double x) { if(fabs(x)<eps) return 0; if(x<0) return -1; return 1; } bool cmp(Point A, Point B) { return A.y<B.y || (A.y==B.y && A.x<B.x); } void Graham(int n) { sort(vertex,vertex+n,cmp); for(int i=0; i<n-1; i++) if(sgn(dist(vertex[i]-vertex[i+1]))==0) pd[vertex[i].res]=pd[vertex[i+1].res]=0; top=0; for(int i=0; i<n; i++) { while(top>1 && (sgn(dist(vertex[Stack[top-1]]-vertex[Stack[top-2]]))==0 || sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-1]]))<0)) vis[Stack[--top]]=0; Stack[top++]=i; vis[i]=1; } int tmp=top; for(int i=n-2; i>=0; i--) { while(top>tmp && (sgn(dist(vertex[Stack[top-1]]-vertex[Stack[top-2]]))==0 || sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-1]]))<0)) vis[Stack[--top]]=0; if(!vis[i]) Stack[top++]=i; } //if(n>1) top--; } int main() { while(scanf("%d",&N)!=EOF && N) { int maxv=-1; for(int i=0; i<N; i++) { scanf("%d%d%d",&x,&y,&v); if(v<maxv) continue; if(v>maxv) { maxv=v; n=0; } vertex[n].x=x; vertex[n].y=y; vertex[n].res=i; n++; } memset(vis,0,sizeof(vis)); memset(pd,1,sizeof(pd)); Graham(n); // for(int i=0; i<top; i++) // printf("%f %f ",vertex[Stack[i]].x,vertex[Stack[i]].y); printf("Case #%d: ",++Case); memset(vis,0,sizeof(vis)); if(maxv>0) { for(int i=0; i<top; i++) vis[vertex[Stack[i]].res]=1; } for(int i=0; i<N; i++) printf("%d",vis[i]&&pd[i]); printf(" "); } return 0; }