POJ1556 The Doors 叉积+最短路

题目大意：求从(0,5)到（10,5）的最短距离，起点与终点之间有n堵墙，每个墙有2个门。

题目思路：判断两点间是否有墙（判断两点的连线是否与某一堵墙的线段相交），建立一个图，然后最短路求出就可以了。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define MAX 1005

double Map[MAX][MAX],dist[MAX];
int n,vis[MAX],G[MAX][MAX];//G储存点的序号
struct node
{
    double x[],y[];
    int len;
}point[MAX];

double Cross(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)//叉积
{
    double a=(x2-x1)*(y3-y1)-(x3-x1)*(y2-y1);
    double b=(x2-x1)*(y4-y1)-(x4-x1)*(y2-y1);
    return a*b;
}

int check(double x1,double y1,double x2,double y2,int pos1,int pos2)//判断两点的连线是否与某一段墙相交
{
    for(int i=pos1+;i<pos2;i++)
    {
        if(Cross(x1,y1,x2,y2,point[i].x[],,point[i].x[],point[i].y[])<1e- && Cross(point[i].x[],,point[i].x[],point[i].y[],x1,y1,x2,y2)<1e-)
            return ;
        if(Cross(x1,y1,x2,y2,point[i].x[],point[i].y[],point[i].x[],point[i].y[])<1e- && Cross(point[i].x[],point[i].y[],point[i].x[],point[i].y[],x1,y1,x2,y2)<1e-)
            return ;
        if(Cross(x1,y1,x2,y2,point[i].x[],point[i].y[],point[i].x[],)<1e- && Cross(point[i].x[],point[i].y[],point[i].x[],,x1,y1,x2,y2)<1e-)
            return ;
    }
    return ;
}

double Dist(double x1,double y1,double x2,double y2)//求两点间距离
{
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}

double dij()//最短路
{
    int k;
    double minn;
    memset(vis,,sizeof(vis));
    for(int i=;i<=*n+;i++)
        dist[i]=Map[][i];
    dist[]=;
    vis[]=;
    for(int i=;i<*n+;i++)
    {
        minn=INF;
        for(int j=;j<=*n+;j++)
        {
            if(minn>dist[j] && !vis[j])
            {
                k=j;
                minn=dist[j];
            }
        }
        vis[k]=;
        for(int j=;j<=*n+;j++)
        {
            if(dist[j] > Map[k][j]+dist[k])
                dist[j]=Map[k][j]+dist[k];
        }
    }
    return dist[*n+];
}

int main()
{
    int cnt;
    double x,y;
    while(scanf("%d",&n),n!=-)
    {
        cnt=;
        point[].len=;
        point[n+].len=;
        G[][]=;
        point[].x[]=;
        point[].y[]=;
        point[n+].x[]=;
        point[n+].y[]=;
        G[n+][]=*n+;
        for(int i=;i<MAX;i++)
        for(int j=;j<MAX;j++)
        Map[i][j]=INF;
        for(int i=;i<=n;i++)
        {
            scanf("%lf",&x);
            point[i].len=;
            for(int j=;j<=;j++)
            {
                scanf("%lf",&y);
                point[i].x[j]=x;
                point[i].y[j]=y;
                G[i][j]=++cnt;
            }
        }
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=point[i].len;j++)
            {
                x=point[i].x[j];
                y=point[i].y[j];
                for(int q=i+;q<=n+;q++)
                {
                    for(int f=;f<=point[q].len;f++)
                    {
                        double x1=point[q].x[f];
                        double y1=point[q].y[f];
                        int op=check(x,y,x1,y1,i,q);
                        if(op)
                        {
                            int a=G[i][j];
                            int b=G[q][f];
                            Map[a][b]=Dist(x,y,x1,y1);
                        }
                    }
                }
            }
        }
        double ans=dij();
        printf("%.2lf\n",ans);
    }
    return ;
}