zoukankan      html  css  js  c++  java
  • UVA11853-Paintball(对偶图)

    Problem UVA11853-Paintball

    Accept:229  Submit:1830

    Time Limit: 3000 mSec

    Problem Description

    You are playing paintball on a 1000×1000 square field. A number of your opponents are on the field hiding behind trees at various positions. Each opponent can fire a paintball a certain distance in any direction. Can you cross the field without being hit by a paintball? Assume that the southwest corner of the field is at (0,0) and the northwest corner at (0,1000).

     Input

    The input contains several scenario. Each scenario consists of a line containing n ≤ 1000, the number of opponents. A line follows for each opponent, containing three real numbers: the (x,y) location of the opponent and its firing range. The opponent can hit you with a paintball if you ever pass within his firing range. You must enter the field somewhere between the southwest and northwest corner and must leave somewhere between the southeast and northeast corners.

     Output

    For each scenario, if you can complete the trip, output four real numbers with two digits after the decimal place, the coordinates at which you may enter and leave the field, separated by spaces. If you can enter and leave at several places, give the most northerly. If there is no such pair of positions, print the line:‘IMPOSSIBLE’

     Sample Input

    3
    500 500 499
    0 0 999
    1000 1000 200
     
     

     Sample Ouput

    0.00 1000.00 1000.00 800.00

    题解:这个题思路比较奇特,有了对偶图的思路,这个题就基本上是个水题了。题目问的时能过去,反过来考虑,怎样不能过去。把正方形区域想象成湖,各个士兵形成的攻击范围看作一个个踏板,如果能够从上边界走到下边界,那么整个图就被分成了左右两部分,自然不能从左到右。并且这个条件时充要的,因此可以作为判据。之后就是如何找最北边。只看左边,如果某个踏板从上边界的踏板出发不能够到达,那该踏板就不对入口坐标造成影响,反过来,如果能到达,并且该踏板和左边界右交点,那么沿着它的上交点就能走回上边界,相当于左上角这一块被包围了,自然入口在下面,据此得到结论,只需要找出从上边界出发能到达的圆,计算出这些圆和左边界交点的最小值就是最北的入口,右边界同理。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <cmath>
     6 using namespace std;
     7 const int maxn = 1000+5;
     8 const double wide = 1000.0;
     9 
    10 struct Point{
    11     double x,y,r;
    12 }point[maxn];
    13 
    14 bool intersection(Point &a,Point &b){
    15     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)) <= a.r+b.r;
    16 }
    17 
    18 int n;
    19 bool vis[maxn];
    20 double Left,Right;
    21 
    22 void check_circle(int a){
    23     if(point[a].x-point[a].r < 0) Left = min(Left,point[a].y-sqrt(point[a].r*point[a].r-point[a].x*point[a].x));
    24     if(point[a].x+point[a].r > wide) Right = min(Right,point[a].y-sqrt(point[a].r*point[a].r-(wide-point[a].x)*(wide-point[a].x)));
    25 }
    26 
    27 bool dfs(int u){
    28     if(vis[u]) return false;
    29     vis[u] = true;
    30     if(point[u].y-point[u].r < 0) return true;
    31     for(int v = 1;v <= n;v++){
    32         if(v==u || vis[v]) continue;
    33         if(intersection(point[u],point[v]) && dfs(v)) return true;
    34     }
    35     check_circle(u);
    36     return false;
    37 }
    38 
    39 int main()
    40 {
    41     //freopen("input.txt","r",stdin);
    42     while(~scanf("%d",&n)){
    43         memset(vis,false,sizeof(vis));
    44         Left = Right = wide;
    45         for(int i = 1;i <= n;i++){
    46             scanf("%lf%lf%lf",&point[i].x,&point[i].y,&point[i].r);
    47         }
    48         bool ok = true;
    49         for(int i = 1;i <= n;i++){
    50             if(!vis[i] && point[i].y+point[i].r>=wide && dfs(i)){
    51                 ok = false;
    52                 break;
    53             }
    54         }
    55         if(ok) printf("%.2f %.2f %.2f %.2f
    ",0.000,Left,wide,Right);
    56         else printf("IMPOSSIBLE
    ");
    57     }
    58     return 0;
    59 }
  • 相关阅读:
    关于vue2.x使用axios以及http-proxy-middleware代理处理跨域的问题
    vue-resource的使用
    从头开始开发一个vue幻灯片组件
    图与例解读Async/Await
    浅谈web缓存
    APICloud框架——总结一下最近开发APP遇到的一些问题 (三)
    编写现代 CSS 代码的 20 个建议
    仿微信联系人列表滑动字母索引
    初来乍到,向各位大牛虚心学习
    转发80端口的脚本
  • 原文地址:https://www.cnblogs.com/npugen/p/9520797.html
Copyright © 2011-2022 走看看