题目链接:http://poj.org/problem?id=1556

Time Limit: 1000MS Memory Limit: 10000K

Description

You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length. 

Input

The input data for the illustrated chamber would appear as follows.


4 2 7 8 9 
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.

Output

The output should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.

Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output

10.00
10.06

题意:

给出一个(0,0)(0,10)(10,0)(10,10)的正方形房子,里面有一些墙,每堵墙上有两扇门;

求从(0,5)到(10,5)的最短距离;

题解:

是一道不错的题目,一定程度上结合了计算几何和最短路;

建一个有向图,把(0,5)作为起始点,(10,5)作为目标点,其他所有墙上的门的两个端点也加入到这个有向图中;

尝试枚举连接任意两个点,只有当这两个点之间没有墙阻隔,这连个点才可能连接起来;

所有能连接起来的两个点都作为一条有向边加入到有向图中,并且靠左的作为u,靠右的作为v,而两点之间的距离作为w;

最后,我们只要求出起始点和目标点之间的最短路即可。

AC代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>
#define MAXN 4*20
#define INF 0x3f3f3f3f
using namespace std; //--------------------------------------计算几何模板 - st-------------------------------------- const double eps = 1e-; struct Point{
double x,y;
Point(double tx=,double ty=):x(tx),y(ty){}
};
typedef Point Vctor; //向量的加减乘除
Vctor operator + (Vctor A,Vctor B){return Vctor(A.x+B.x,A.y+B.y);}
Vctor operator - (Point A,Point B){return Vctor(A.x-B.x,A.y-B.y);}
Vctor operator * (Vctor A,double p){return Vctor(A.x*p,A.y*p);}
Vctor operator / (Vctor A,double p){return Vctor(A.x/p,A.y/p);} int dcmp(double x)
{
if(fabs(x)<eps) return ;
else return (x<)?(-):();
}
bool operator == (Point A,Point B){return dcmp(A.x-B.x)== && dcmp(A.y-B.y)==;} //向量的点积,长度,夹角
double Dot(Vctor A,Vctor B){return A.x*B.x+A.y*B.y;}
double Length(Vctor A){return sqrt(Dot(A,A));}
double Angle(Vctor A,Vctor B){return acos(Dot(A,B)/Length(A)/Length(B));} //叉积,三角形面积
double Cross(Vctor A,Vctor B){return A.x*B.y-A.y*B.x;}
double TriangleArea(Point A,Point B,Point C){return Cross(B-A,C-A);} //判断线段是否规范相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1,b2 - a1),
c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1,a2 - b1);
return dcmp(c1)*dcmp(c2)< && dcmp(c3)*dcmp(c4)<;
} //--------------------------------------计算几何模板 - ed-------------------------------------- //--------------------------------------spfa算法 - st--------------------------------------
double d[MAXN];
double mp[MAXN][MAXN];
bool vis[MAXN];
void init(int n){for(int i=;i<n;i++) for(int j=;j<n;j++) mp[i][j]=;}
void addedge(int u,int v,double w){mp[u][v]=w;}
void spfa(int st,int n)
{
for(int i=;i<n;i++)
{
i==st ? d[i]= : d[i]=INF;
vis[i]=;
}
queue<int> q;
q.push(st);
vis[st]=;
while(!q.empty())
{
int u=q.front();q.pop();vis[u]=;
for(int v=;v<n;v++)
{
if(u==v || mp[u][v]==) continue;
double tmp=d[v];
if(d[v]>d[u]+mp[u][v]) d[v]=d[u]+mp[u][v];
if(d[v]<tmp && !vis[v])
{
q.push(v);
vis[v]=;
}
}
}
}
//--------------------------------------spfa算法 - ed-------------------------------------- int n;
vector<Point> p;
int main()
{
while(scanf("%d",&n) && n!=-)
{
p.clear();
p.push_back(Point(,));
for(int i=;i<=n;i++)
{
double x,y1,y2,y3,y4;
scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
p.push_back(Point(x,y1));
p.push_back(Point(x,y2));
p.push_back(Point(x,y3));
p.push_back(Point(x,y4));
}
p.push_back(Point(,)); int _size=p.size();
init(_size);
for(int i=;i<_size;i++)
{
for(int j=i+;j<_size;j++)
{
if(p[i].x==p[j].x) continue; bool ok=;
for(int k=i+;k<j;k++)
{
if(k%==)
{
if(SegmentProperIntersection(p[i],p[j],p[k],Point(p[k].x,)))
{
ok=;
break;
}
}
else if(k%==)
{
if(SegmentProperIntersection(p[i],p[j],p[k],p[k+]))
{
ok=;
break;
}
}
else if(k%==)
{
if(SegmentProperIntersection(p[i],p[j],p[k],p[k-]))
{
ok=;
break;
}
}
else if(k%==)
{
if(SegmentProperIntersection(p[i],p[j],p[k],Point(p[k].x,)))
{
ok=;
break;
}
}
}
if(ok) addedge(i,j,Length(p[i]-p[j]));
}
} spfa(,_size);
printf("%.2f\n",d[_size-]);
}
}

POJ 1556 - The Doors - [平面几何+建图spfa最短路]的更多相关文章

  1. NOIP2013 华容道 (棋盘建图+spfa最短路)

    #include <cstdio> #include <algorithm> #include <cstring> #include <queue> # ...

  2. poj 1556 The Doors(线段相交,最短路)

      The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7430   Accepted: 2915 Descr ...

  3. POJ 1556 The Doors 线段交 dijkstra

    LINK 题意:在$10*10$的几何平面内,给出n条垂直x轴的线,且在线上开了两个口,起点为$(0, 5)$,终点为$(10, 5)$,问起点到终点不与其他线段相交的情况下的最小距离. 思路:将每个 ...

  4. POJ 1556 - The Doors 线段相交不含端点

    POJ 1556 - The Doors题意:    在 10x10 的空间里有很多垂直的墙,不能穿墙,问你从(0,5) 到 (10,5)的最短距离是多少.    分析:        要么直达,要么 ...

  5. poj 2135 Farm Tour 最小费用最大流建图跑最短路

    题目链接 题意:无向图有N(N <= 1000)个节点,M(M <= 10000)条边:从节点1走到节点N再从N走回来,图中不能走同一条边,且图中可能出现重边,问最短距离之和为多少? 思路 ...

  6. POJ 3687 Labeling Balls 逆向建图,拓扑排序

    题目链接: http://poj.org/problem?id=3687 要逆向建图,输入的时候要判重边,找入度为0的点的时候要从大到小循环,尽量让编号大的先入栈,输出的时候注意按编号的顺序输出重量, ...

  7. Invitation Cards(邻接表+逆向建图+SPFA)

    Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 17538   Accepted: 5721 Description In ...

  8. CF786B Legacy 线段树优化建图 + spfa

    CodeForces 786B Rick和他的同事们做出了一种新的带放射性的婴儿食品(???根据图片和原文的确如此...),与此同时很多坏人正追赶着他们.因此Rick想在坏人们捉到他之前把他的遗产留给 ...

  9. 简单几何(线段相交+最短路) POJ 1556 The Doors

    题目传送门 题意:从(0, 5)走到(10, 5),中间有一些门,走的路是直线,问最短的距离 分析:关键是建图,可以保存所有的点,两点连通的条件是线段和中间的线段都不相交,建立有向图,然后用Dijks ...

随机推荐

  1. jquery获取元素颜色css('color')的值返回RGB

    css代码如下: a, a:link, a:visited { color:#4188FB; } a:active, a:focus, a:hover { color:#FFCC00; } js代码如 ...

  2. Apache性能优化总结

    1.介绍 首先要了解Apache采用的MPM(Multi -Processing Modules,多道处理模块),MPM是Apache的核心,它的作用是管理网络连接.调度请求.Apache2.0中MP ...

  3. Java -- 异常的捕获及处理 -- throws与throw关键字

    7.2 throws 与 throw关键字 7.2.1 throws 关键字 在定义一个方法时可以使用throws关键字声明,使用throws声明的方法标识此方法不处理异常,而交给方法的调用处进行处理 ...

  4. Android实现图片缩放示例

    package com.example.demo; import android.os.Bundle; import android.app.Activity; import android.grap ...

  5. python --help查询python相关命令

    C:\Users\lenovo>python --help usage: python [option] ... [-c cmd | -m mod | file | -] [arg] ... O ...

  6. IIS中采用ISAPI-Rewrite防盗链

    本规则支持白名单排除式防盗链,搜索引擎友好(不屏蔽),被盗链后的错误提示转向,支持各种文件类型,经作者亲验真的能用,第一时间在itmop.com原创发表,请继续往下阅读. 近来小站遇到了盗链问题,至使 ...

  7. centos6.4搭建ftp服务器

    在centos环境下搭建ftp服务器,选择用vsftpd. 1.检测是否已经安装vsftpd # rpm -qa | grep vsftpd 如果已经安装vsftpd,会显示相应版本号.没有任何提示, ...

  8. js 中的break continue return

    break:跳出整个循环 1.当i=6时,就跳出了整个循环,此for循环就不继续了: continue:跳出当前循环,继续下一次循环: return :指定函数返回值 1.在js当中,常使用retur ...

  9. codeforces水题100道 第一题 Codeforces Beta Round #1 A. Theatre Square (math)

    题目链接:http://www.codeforces.com/problemset/problem/1/A题意:至少用多少块边长为a的方块铺满NxM的矩形区域.C++代码: #include < ...

  10. Android学习之蓝牙操作

    BluetoothAdapter 用法 蓝牙运行原理:通过BluetoothAdapter 蓝牙适配器处理任务,如果蓝牙被启动之后,系统会自动去搜索其它设备,如果匹配到附近的设备就发送一个广播,Bro ...