zoukankan      html  css  js  c++  java
  • [POJ 2588] Snakes

    同swustoj 8

    Snakes
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 1015   Accepted: 341

    Description

    Buffalo Bill wishes to cross a 1000x1000 square field. A number of snakes are on the field at various positions, and each snake can strike a particular distance in any direction. Can Bill make the trip without being bitten?

    Input

    Assume that the southwest corner of the field is at (0,0) and the northwest corner at (0,1000). The input consists of a line containing n <= 1000, the number of snakes. A line follows for each snake, containing three real numbers: the (x,y) location of the snake and its strike distance. The snake will bite anything that passes closer than this distance from its location.

    Output

    Bill must enter the field somewhere between the southwest and northwest corner and must leave somewhere between the southeast and northeast corners. 

    If Bill can complete the trip, give coordinates at which he may enter and leave the field. If Bill may enter and leave at several places, give the most northerly. If there is no such pair of positions, print "Bill will be bitten." 

    Sample Input

    3
    500 500 499
    0 0 999
    1000 1000 200
    

    Sample Output

    Bill enters at (0.00, 1000.00) and leaves at (1000.00, 800.00).
    

    并查集+计算几何

    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<vector>
    #include<cstdio>
    using namespace std;
    
    #define PI acos(-1.0)
    #define EPS 1e-8
    #define N 1010
    
    int dcmp(double x)
    {
        if(fabs(x)<EPS) return 0;
        return x<0?-1:1;
    }
    struct Point
    {
        double x,y;
        Point (){}
        Point (double x,double y):x(x),y(y){}
        Point operator - (Point p){
            return Point(x-p.x,y-p.y);
        }
        bool operator == (Point p){
            return dcmp(fabs(x-p.x))==0 && dcmp(fabs(y-p.y))==0;
        }
        double operator * (Point p){
            return x*p.x+y*p.y;
        }
        double operator ^ (Point p){
            return x*p.y-y*p.x;
        }
        double length(){
            return sqrt(x*x+y*y);
        }
        double angle(){
            return atan2(y,x);
        }
        bool operator <(const Point &p)const{
            return y<p.y;
        }
    };
    struct Line
    {
        Point s,e;
        Line (){}
        Line (Point s,Point e):s(s),e(e){}
        Point GetPoint(double t){
            return Point(s.x+(e.x-s.x)*t,s.y+(e.y-s.y)*t);
        }
    };
    struct Circle
    {
        Point c;
        double r;
        Circle(){}
        Circle(Point c,double r):c(c),r(r){}
        Point GetPoint(double a){
            return Point(c.x+cos(a)*r,c.y+sin(a)*r);
        }
        /* 0表示相离,1表示相切,2表示相交 */
        pair<int,vector<Point> > CircleInterLine(Line l){
            vector<Point> res;
            double A=l.e.x-l.s.x,B=l.s.x-c.x,C=l.e.y-l.s.y,D=l.s.y-c.y;
            double E=A*A+C*C,F=2*(A*B+C*D),G=B*B+D*D-r*r;
            double delta=F*F-4*E*G;
            if(dcmp(delta)<0) return make_pair(0,res);
            if(dcmp(delta)==0){
                res.push_back(l.GetPoint(-F/(2*E)));
                return make_pair(1,res);
            }
            res.push_back(l.GetPoint((-F-sqrt(delta))/(2*E)));
            res.push_back(l.GetPoint((-F+sqrt(delta))/(2*E)));
            return make_pair(2,res);
        }
        /* -1表示重合,0表示相离,1表示相切,2表示相交 */
        int operator & (Circle C){
            double d=(c-C.c).length();
            if(dcmp(d)==0){
                if(dcmp(r-C.r)==0) return -1;
                return 0;
            }
            if(dcmp(r+C.r-d)<0) return 0;
            if(dcmp(fabs(r-C.r)-d)>0) return 0;
            double a=(C.c-c).angle();
            double da=acos((r*r+d*d-C.r*C.r)/(2*r*d));
            Point p1=GetPoint(a-da),p2=GetPoint(a+da);
            if(p1==p2) return 1;
            return 2;
        }
    };
    
    int n;
    int f[N];
    Line up,down;
    Line lft,rgt;
    Circle c[N];
    
    void init()
    {
        for(int i=0;i<=n+1;i++) f[i]=i;
        up=Line(Point(0,1000),Point(1000,1000));
        down=Line(Point(0,0),Point(1000,0));
        lft=Line(Point(0,0),Point(0,1000));
        rgt=Line(Point(1000,0),Point(1000,1000));
    }
    int Find(int x)
    {
        if(x!=f[x]) f[x]=Find(f[x]);
        return f[x];
    }
    void UN(int x,int y)
    {
        x=Find(x);
        y=Find(y);
        if(x!=y) f[x]=y;
    }
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            init();
            for(int i=1;i<=n;i++) scanf("%lf%lf%lf",&c[i].c.x,&c[i].c.y,&c[i].r);
            for(int i=1;i<=n;i++){
                for(int j=i+1;j<=n;j++){
                    if((c[i]&c[j])!=0){
                        UN(i,j);
                    }
                }
            }
            //上边界
            for(int i=1;i<=n;i++){
                pair<int,vector<Point> > res=c[i].CircleInterLine(up);
                if(res.first!=0) UN(0,i);
            }
            //下边界
            for(int i=1;i<=n;i++){
                pair<int,vector<Point> > res=c[i].CircleInterLine(down);
                if(res.first!=0) UN(i,n+1);
            }
            if(Find(0)==Find(n+1)){ //出不去
                printf("Bill will be bitten.
    ");
                continue;
            }
            //左右边界
            vector<Point> p1,p2;
            p1.push_back(Point(0,1000));
            p2.push_back(Point(1000,1000));
            for(int i=1;i<=n;i++){
                pair<int,vector<Point> > res1=c[i].CircleInterLine(lft);
                pair<int,vector<Point> > res2=c[i].CircleInterLine(rgt);
                if(res1.first!=0){
                    while(!res1.second.empty()){
                        if(res1.second.back().y>=0 && res1.second.back().y<=1000 && Find(i)==Find(0))
                            p1.push_back(res1.second.back());
                        res1.second.pop_back();
                    }
                }
                if(res2.first!=0){
                    while(!res2.second.empty()){
                        if(res2.second.back().y>=0 && res2.second.back().y<=1000 && Find(i)==Find(0))
                            p2.push_back(res2.second.back());
                        res2.second.pop_back();
                    }
                }
            }
            int i,j;
            sort(p1.begin(),p1.end());
            sort(p2.begin(),p2.end());
            printf("Bill enters at (0.00, %.2f) and leaves at (1000.00, %.2f).
    ",p1[0].y,p2[0].y);
        }
        return 0;
    }
  • 相关阅读:
    Bootstrap 4/3 页面基础模板 与 兼容旧版本浏览器
    Asp.net core 项目实战 新闻网站+后台 源码、设计原理 、视频教程
    C# 数据类型转换 显式转型、隐式转型、强制转型
    C# 多维数组 交错数组的区别,即 [ , ] 与 [ ][ ]的区别
    C#/Entity Frame Core 使用Linq 进行分页 .Skip() .Take() 的使用方法
    利用 Xunsearch 搭建搜索引擎、内容搜索实战
    Delphi
    Python
    Python
    Python
  • 原文地址:https://www.cnblogs.com/hate13/p/4598163.html
Copyright © 2011-2022 走看看