链接:http://poj.org/problem?id=1556

The Doors
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 6216   Accepted: 2495

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 ////////////////////////////////////////////////////////////////////
这题处理起来挺难的,要把输入的点存到图里,用Dijkstra求出最短路径,存图的过程是,判断任意两点连成的线,横坐标不能相同,并且如果,横坐标与线上的横坐标不相同,就要判断是否相交,相交则行不通
否则存图,用Dijkstra搜出最短的路径即可
还有,要注意细节
 #include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <math.h>
#include <algorithm> #define eps 1e-6
#define INF 1000000000
typedef struct point
{
double x,y;
}point; typedef struct beline
{
point st,ed;
}beline; using namespace std; point p[];
double mp[][];
double d[];
int visit[]; bool dy(double x,double y){ return x > y+eps; }
bool xy(double x,double y){ return x < y-eps; }
bool dyd(double x,double y){ return x > y-eps; }
bool xyd(double x,double y){ return x < y+eps; }
bool dd(double x,double y){ return fabs(x - y)<eps; } double crossProduct(point a,point b,point c)
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
}
double Dist(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
} bool onSegment(point a,point b,point c)
{
double maxx=max(a.x,b.x);
double maxy=max(a.y,b.y);
double minx=min(a.x,b.x);
double miny=min(a.y,b.y);
if(dd(crossProduct(a,b,c),0.0)&&dy(c.x,minx)&&xy(c.x,maxx)
&&dy(c.y,miny)&&xy(c.y,maxy))
return true;
return false;
} bool segIntersect(point p1,point p2,point p3,point p4)
{
double d1 = crossProduct(p3,p4,p1);
double d2 = crossProduct(p3,p4,p2);
double d3 = crossProduct(p1,p2,p3);
double d4 = crossProduct(p1,p2,p4);
if(xy(d1*d2,0.0)&&xy(d3*d4,0.0))
return true;
if(dd(d1,0.0)&&onSegment(p3,p4,p1))
return true;
if(dd(d2,0.0)&&onSegment(p3,p4,p2))
return true;
if(dd(d3,0.0)&&onSegment(p1,p2,p3))
return true;
if(dd(d4,0.0)&&onSegment(p1,p2,p4))
return true;
return false;
} void Dijkstra(int n)
{
int i,y;
memset(visit,,sizeof(visit));
for(i=; i<n; i++)
d[i] = mp[][i];
d[] = ;
for(i=; i<n; i++)
{
int m=INF,x;
{
for(y=; y<n; y++)
{
if(!visit[y] && d[y]<=m)
{
m = d[ x = y ];
}
}
visit[x]=;
for(y=; y<n; y++)
{
if(!visit[y] && d[y] > d[x]+mp[x][y])
{
d[y] = d[x] + mp[x][y];
}
}
}
}
} int main()
{
int n,m,i,j,k,t;
double a,b,c,d1,e;
beline li[];
beline tmp;
p[].x=;p[].y=;//freopen("in.txt","r",stdin);
while(scanf("%d",&n)!=EOF && n!=-)
{
for(i=; i<; i++)
for(j=; j<; j++)
mp[i][j] = INF;
int cas=,css=;
for(i=; i<n; i++)
{
scanf("%lf%lf%lf%lf%lf",&a,&b,&c,&d1,&e);
li[css].st.x=a;
li[css].st.y=;
p[cas].x=a; li[css].ed.x=a;
p[cas++].y=b;li[css++].ed.y=b;
p[cas].x=a; li[css].st.x=a;
p[cas++].y=c;li[css].st.y=c;
p[cas].x=a; li[css].ed.x=a;
p[cas++].y=d1;li[css++].ed.y=d1;
p[cas].x=a; li[css].st.x=a;
p[cas++].y=e;li[css].st.y=e;
li[css].ed.x=a;
li[css++].ed.y=;
}
p[cas].x=10.0;p[cas].y=5.0;
for(i=; i<=cas; i++)
{
for(j=i+; j<=cas; j++)
{
int ok=;
for(k=; k<css; k++)
{
if(dd(p[i].x,p[j].x)||!dd(p[i].x,li[k].st.x)&&!dd(p[j].x,li[k].st.x)&&(segIntersect(p[i],p[j],li[k].st,li[k].ed)))
{
ok=;
break;
}
}
if(!ok)
{
mp[j][i] = mp[i][j] = Dist(p[i],p[j]);//printf("%d %d %lf ^^\n",i,j,mp[i][j]);
}
}
}
Dijkstra(cas+);
printf("%.2lf\n",d[cas]);
}
return ;
}

poj 1556 (Dijkstra + Geometry 线段相交)的更多相关文章

  1. 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 ...

  2. POJ 1556 计算几何 判断线段相交 最短路

    题意: 在一个左下角坐标为(0,0),右上角坐标为(10,10)的矩形内,起点为(0,5),终点为(10,5),中间会有许多扇垂直于x轴的门,求从起点到终点在能走的情况下的最短距离. 分析: 既然是求 ...

  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(线段交+最短路)

    #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm& ...

  5. POJ 1066 Treasure Hunt(线段相交判断)

    Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4797   Accepted: 1998 Des ...

  6. POJ 2653 Pick-up sticks(线段相交)

    题意:给定n个木棍依次放下,要求最终判断没被覆盖的木棍是哪些. 思路:快速排斥以及跨立实验可以判断线段相交. #include<algorithm> #include<cstdio& ...

  7. poj 3304(直线与线段相交)

    传送门:Segments 题意:线段在一个直线上的摄影相交 求求是否存在一条直线,使所有线段到这条直线的投影至少有一个交点 分析:可以在共同投影处作原直线的垂线,则该垂线与所有线段都相交<==& ...

  8. [poj 1127]Jack Straws[线段相交][并查集]

    题意: 给出一系列线段,判断某两个线段是否连通. 思路: 根据线段相交情况建立并查集, 在同一并查集中则连通. (第一反应是强连通分量...实际上只要判断共存即可, 具体的方向啊是没有关系的..) 并 ...

  9. POJ 2653 Pick-up sticks [线段相交 迷之暴力]

    Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 12861   Accepted: 4847 D ...

随机推荐

  1. sql创建删除修改表的基本操作

    1 建立表格 在建立好数据库以后,就可以根据储存资料的需求,使用SQL叙述建立所有需要的表格(table).建立表格的设定非常多,以建立"world.city"表格来说,它的叙述会 ...

  2. 关于 MySQL LEFT JOIN 你可能需要了解的三点

    即使你认为自己已对 MySQL 的 LEFT JOIN 理解深刻,但我敢打赌,这篇文章肯定能让你学会点东西! ON 子句与 WHERE 子句的不同 一种更好地理解带有 WHERE ... IS NUL ...

  3. iOS 学习笔记 七 (2015.03.29)code snippet操作

    1.code snippet 备份路径:~/Library/Developer/Xcode/UserData/CodeSnippets/

  4. sersync实现触发式同步

    金山的一个居于inotify+rsync进行二次开发实现文件同步的小工具sersync,能够很方便的实现文件触发式同步 Inotify 是基于inode级别的文件系统监控技术,是一种强大的.细粒度的. ...

  5. jQuery中attr() 和 prop()【转】

    Version 1.9.0 开始不建议使用 attr() 来对具有 true 和 false 两个属性的属性进行操作. 具有 true 和 false 两个属性的属性,如 checked, selec ...

  6. Java获取字符串编码方式

    直接下载吧: http://files.cnblogs.com/files/xiluhua/BytesEncodingDetectTool.rar

  7. WPF:窗体置顶

    1.设置窗体TopMost属性 private DispatcherTimer timer; public Window1() { InitializeComponent(); Loaded += n ...

  8. Android Studio解决unspecified on project app resolves to an APK archive which is not supported

    出现该问题unspecified on project app resolves to an APK archive which is not supported as a compilation d ...

  9. JAVA字段的初始化规律

    JAVA字段的初始化规律 1.类的构造方法 (1)“构造方法”,也称为“构造函数”,当创建一个对象时,它的构造方法会被自动调用.构造方法与类名相同,没有返回值. (2)如果类没有定义构造函数,Java ...

  10. Android动画的使用总结

    1.补间动画(透明渐变.平移.旋转.缩放.组合) 方法一:通过xml文件设置 1-1:创建:res/anim 1-2:java代码写调用 Animation a = AnimationUtils.lo ...