HDU 4946 共线凸包

 

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("\n");
    }
    return 0;
}

/*
0 0 1
0 1
0 1
1 1
2 1
3 1
1 1

*/
View Code

 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("\n");
    }
    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

*/
View Code

 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\n",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("\n");
    }
    return 0;
}
View Code

 

posted @ 2014-08-17 13:54  Estimator  阅读(459)  评论(0编辑  收藏  举报