POJ1556 最短路 + 线段相交问题

POJ1556

题目大意：比较明显的题目，在一个房间中有几堵墙，直着走，问你从（0，5）到（10，5）的最短路是多少

求最短路问题，唯一变化的就是边的获取，需要我们获取边，这就需要判断我们想要走的这条边会不会经过墙

所以创建点集，线段集合

#include <iostream>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <iomanip>
#define eps 1e-10
#define inf 0x3f3f3f3f
#define equal(a,b) fabs((a) - (b)) < eps
using namespace std;
const int maxn = 200;
int  pnum,lnum;
struct Point
{
    double x,y;
    Point(double x = 0.0,double y = 0.0):x(x),y(y){}

    Point operator - (Point p){return Point(x-p.x,y-p.y);}

}ps[maxn];
struct segment
{
    Point p1,p2;
    segment(Point p1 = Point(0.0,0.0),Point p2 = Point(0.0,0.0)):p1(p1),p2(p2){}
}ls[maxn];
//存储线段和点~~

 事先也要准备好最短路的东西

double mp[maxn][maxn];
double dis[maxn];
int vis[maxn];

 

int main()
{
    int n;
    double x,y1,y2,y3,y4;
    while(~scanf("%d",&n),n != -1)
    {
        init();
        ps[pnum++] = Point(0,5);
        for(int i = 0;i < n;i++)
        {
            scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
            ps[pnum++] = Point(x,y1);
            ps[pnum++] = Point(x,y2);
            ps[pnum++] = Point(x,y3);
            ps[pnum++] = Point(x,y4);
            ls[lnum++] = segment(Point(x,0),Point(x,y1));
            ls[lnum++] = segment(Point(x,y2),Point(x,y3));
            ls[lnum++] = segment(Point(x,y4),Point(x,10));
        }
        ps[pnum] = Point(10,5);
        for(int i = 0;i <= pnum;i++)
        {
            for(int j = 0;j <= pnum;j++)
            {
                if(i == j)mp[i][j] = 0.0;
                else if(checkline(i,j))
                {
                    mp[i][j] = getdis(i,j);
                    //printf("%d   %d\n",i,j);
                    //cout<<mp[i][j]<<endl;
                    //cout<<mp[i][j]<<endl;
                }
                else mp[i][j] = inf * 1.0;
            }
        }
        dijkscar(0,pnum);
        cout << fixed << setprecision(2) << dis[pnum] << endl;

    }
    return 0;
}

 main函数中是输入点和线段至对应的集合中，然后去找合适的边填充mp二维数组，这是侯用到了线段相交的判断

bool intersect(int i,int j,int k)
{
    if(cross(ps[i],ps[j],ls[k].p1) * cross(ps[i],ps[j],ls[k].p2) < -eps && 
cross(ls[k].p1,ls[k].p2,ps[i]) * cross(ls[k].p1,ls[k].p2,ps[j]) < -eps)return true;
    return false;
}
bool checkline(int i,int j)
{
    for(int k = 0;k < lnum;k++)
    {
        //printf("k = %d && i = %d && j = %d  Point[i] = (%d,%d),Point[j] = (%d,%d),Segment = %d %d   %d %d\n",k,i,
        //       j,(int)ps[i].x,(int)ps[i].y,(int)ps[j].x,(int)ps[j].y,(int)ls[k].p1.x,(int)ls[k].p1.y,(int)ls[k].p2.x,(int)ls[k].p2.y);
        if(intersect(i,j,k))
            return false;
    }
    return true;
}
double cross(Point p1,Point p2,Point p3)
{
    Point a = p2 - p1;
    Point b = p3 - p1;
    return a.x * b.y - a.y * b.x;
}

 线段是否相交，利用外积的方向判断，如果双方都符合一条线段的两个端点在另一条线段的两个端点时，就会相交

然后时常规的dijkscar算法

void dijkscar(int s,int n)
{
    for(int i = 1;i <= n;i++)dis[i] = mp[s][i];
    dis[s] = 0;
    vis[s] = 1;
    for(int i = 0;i <= n;i++)//优化次数
    {
        double minlen = inf * 1.0;
        int net = 0;
        for(int j = 1;j <= n;j++)
        {
            if(!vis[j] && dis[j] < minlen)
            {
                minlen = dis[j];
                net = j;
            }
            //if(net == 0)break;
            //if(minlen == inf * 1.0)break;
            vis[net] = 1;
            for(int j = 1;j <= n;j++)
            {
                if(dis[j] > dis[net] + mp[net][j])
                {
                    dis[j] = dis[net] + mp[net][j];
                }
            }
        }
    }

}

 我犯的几个错误

dijkscar中if(net == 0)break;这句话导致我最短路判断失误，我目前也不知道什么原因

符合条件的边输入进mp数组，我以为只要输入单向向右的就行了所以一开始的for循环时i = 0 -> n - 1 + j = i + 1 - > n，结果发现，emm还是有特殊情况的，比如径直往上走的情况~~