[日常摸鱼]UVA393 The Doors 简单计算几何+最短路

The  Boy Next   Doors

题意：给定一个固定大小的房间（$x,y$的范围都是$[0,10]$），有$n$个墙壁作为障碍（都与横坐标轴垂直），每个墙壁都有两扇门分别用四个点来描述，起点终点固定在$(0,5)$和$(10,5)$，求起点到终点的最短路长度，$n<=18$

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题解：

我们把每堵墙的每一“段”作为一条线段，对任意两点$u,v$，如果两点间的连线不和其他线段相交，那我们从$u$走到$v$的最短距离就是他们的欧几里得距离，对所有点对都这么做一遍，处理出所有能够直接到达的点之间的距离

如果不能直接从$u$到$v$，那么既然是最短路，那走直线一定比走曲线要好，也就是一定是经过他们中间的某些点然后到达的，然后对所有点跑一次最短路就好了

这题范围很小Floyd也能过…

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

const double precision=10e-6;

const double INF=(~0u>>2);

const int N=105;

struct Point 
{
    double x,y;
}p[N];

struct Line
{
    Point a,b;
}l[N];

int n,cnt_point,cnt_line;

double dist[N][N];

inline int dblcmp(double d)
{
    if(fabs(d)<precision)
        return 0;
    return (d>0?1:-1);
}

inline double det(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}

inline double cross(Point a,Point b,Point c)
{
    return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y);
}

inline bool SegCrossSimple(Point a,Point b,Point c,Point d)
{
    return (dblcmp(cross(c,a,b))^dblcmp(cross(d,a,b)))==-2&&(dblcmp(cross(a,c,d))^dblcmp(cross(b,c,d)))==-2;
}

inline double sqr2(double x)
{
    return x*x;
}

inline double dis(int i,int j)
{
    return sqrt(sqr2(p[i].x-p[j].x)+sqr2(p[i].y-p[j].y));
}

inline void addPoint(double x,double y)
{
    p[++cnt_point]=(Point){x,y};
}

inline void addLine(double x1,double y1,double x2,double y2)
{
    l[++cnt_line]=(Line){(Point){x1,y1},(Point){x2,y2}};
}

inline void init()
{
    memset(p,0,sizeof(p));
    memset(l,0,sizeof(l));
    cnt_point=cnt_line=0;
    addPoint(0,5);addPoint(10,5);
}

inline void solve()
{
    init();
    
    double x,y1,y2,y3,y4;
    for(register int i=1;i<=n;i++)
    {
        scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
        addPoint(x,y1);addPoint(x,y2);
        addPoint(x,y3);addPoint(x,y4);
        addLine(x,0,x,y1);addLine(x,y2,x,y3);addLine(x,y4,x,10);
    }
    
    for(register int i=1;i<=cnt_point;i++)dist[i][i]=0;
    
    for(register int i=1;i<=cnt_point;i++)
    {
        for(register int j=i+1;j<=cnt_point;j++)
        {
            bool flag=1;
            for(register int k=1;k<=cnt_line;k++)
            {
                if(SegCrossSimple(p[i],p[j],l[k].a,l[k].b))
                {
                    flag=0;
                    break;
                }
            }
            if(flag)
                dist[i][j]=dis(i,j);
            else
                dist[i][j]=INF;
                
            dist[j][i]=dist[i][j];
        }
    }
        
    for(register int k=1;k<=cnt_point;k++)
        for(register int i=1;i<=cnt_point;i++)
            for(register int j=1;j<=cnt_point;j++)if(dist[i][k]!=INF&&dist[k][j]!=INF)
                dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
                
    printf("%.2lf\n",dist[1][2]);
}

int main()
{
    while(scanf("%d",&n)==1)
    {
        if(n==-1)break;
        solve();
    }
    return 0;
}