POJ (线段相交 最短路) The Doors

题意：

一个正方形中有n道竖直的墙，每道墙上开两个门。求从左边中点走到右边中点的最短距离。

分析：

以起点终点和每个门的两个端点建图，如果两个点可以直接相连(即不会被墙挡住)，则权值为两点间的欧几里得距离。

然后求起点到终点的最短路即可。

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

const int maxn = 50;
const double INF = 1e4;
const double eps = 1e-8;

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x), y(y) {}
}p[maxn * 4];

typedef Point Vector;

Point read_point()
{
    double x, y;
    scanf("%lf%lf", &x, &y);
    return Point(x, y);
}

Point operator - (const Point& A, const Point& B)
{ return Point(A.x-B.x, A.y-B.y); }

Vector operator / (const Vector& A, double p)
{ return Vector(A.x/p, A.y/p); }

double Dot(const Vector& A, const Vector& B)
{ return A.x*B.x + A.y*B.y; }

double Length(const Vector& A)
{ return sqrt(Dot(A, A)); }

struct Door
{
    double x, y1, y2, y3, y4;
    Door(double x=0, double y1=0, double y2=0, double y3=0, double y4=0):x(x), y1(y1), y2(y2), y3(y3), y4(y4) {}
};

vector<Door> door;

double d[maxn * 4], w[maxn * 4][maxn * 4];
bool vis[maxn * 4];
int cnt;

bool isOK(int a, int b)
{//判断两点是否能直接到达
    if(p[a].x >= p[b].x) swap(a, b);
    for(int i = 0; i < door.size(); ++i)
    {
        if(door[i].x <= p[a].x) continue;
        if(door[i].x >= p[b].x) break;
        double k = (p[b].y-p[a].y) / (p[b].x-p[a].x);
        double y = p[a].y + k * (door[i].x - p[a].x);
        if(!(y>=door[i].y1&&y<=door[i].y2 || y>=door[i].y3&&y<=door[i].y4)) return false;
    }
    return true;
}

void Init()
{
    for(int i = 0; i < cnt; ++i)
        for(int j = i; j < cnt; ++j)
            if(i == j) w[i][j] = 0;
            else w[i][j] = w[j][i] = INF;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    int n;
    while(scanf("%d", &n) == 1 && n + 1)
    {
        door.clear();
        memset(vis, false, sizeof(vis));
        memset(d, 0, sizeof(d));

        p[0] = Point(0, 5);
        cnt = 1;
        for(int i = 0; i < n; ++i)
        {
            double x, y[4];
            scanf("%lf", &x);
            for(int j = 0; j < 4; ++j) { scanf("%lf", &y[j]); p[cnt++] = Point(x, y[j]); }
            door.push_back(Door(x, y[0], y[1], y[2], y[3]));
        }
        p[cnt++] = Point(10, 5);

        Init();

        for(int i = 0; i < cnt; ++i)
            for(int j = i+1; j < cnt; ++j)
            {
                double l = Length(Vector(p[i]-p[j]));
                if(p[i].x == p[j].x) continue;
                if(isOK(i, j))
                    w[i][j] = w[j][i] = l;
            }
        //Dijkstra
        d[0] = 0;
        for(int i = 1; i < cnt; ++i) d[i] = INF;
        for(int i = 0; i < cnt; ++i)
        {
            int x;
            double m = INF;
            for(int y = 0; y < cnt; ++y) if(!vis[y] && d[y] <= m) m = d[x=y];
            vis[x] = 1;
            for(int y = 0; y < cnt; ++y) d[y] = min(d[y], d[x] + w[x][y]);
        }

        printf("%.2f
", d[cnt-1]);
    }

    return 0;
}