The Doors - POJ 1556 (线段相交)

题目大意：有一个房间（左上角（0，10），右下角（10,0）），然后房间里有N面墙，每面墙上都有两个门，求出来从初始点（0,5），到达终点（10，5）的最短距离。

分析：很明显根据两点之间直线最短，所以所走的路线一定是点之间的连线，只需要判断一下这两点间知否有墙即可。

代码如下：

======================================================================================================================================

#include<math.h>
#include<algorithm>
#include<stdio.h>
using namespace std;

const int MAXN = 1007;
const double oo = 1e9+7;
const double EPS = 1e-8;

struct point
{
    double x, y, len;
    point(double x=0, double y=0, double len=0):x(x),y(y),len(len){}
    point operator - (const point &t) const{
        return point(x-t.x, y-t.y);
    }
    int operator *(const point &t) const{
        double c = x*t.y - y*t.x;
        if(c > EPS)return 1;
        if(fabs(c)<EPS)return 0;
        return -1;
    }
};
struct Wall
{
    point A, B;
    Wall(point A=0, point B=0):A(A), B(B){}
};

bool CanLine(point a, point b, Wall w[], int N)
{
    for(int i=0; i<N; i++)
    {
        if( w[i].A.x < b.x || w[i].A.x > a.x )
            continue;
        int t = (a-b)*(w[i].A-b) + (a-b)*(w[i].B-b);

        if(t == 0)
            return false;
    }

    return true;
}

int main()
{
    int M;

    while(scanf("%d", &M) != EOF && M != -1)
    {
        int i, j, nw=0, np=1;
        double x, y[10];
        Wall w[MAXN]; point p[MAXN];

        p[0] = point(0, 5, 0);
        while(M--)
        {
            scanf("%lf%lf%lf%lf%lf", &x, &y[0], &y[1], &y[2], &y[3]);

            p[np++] = point(x, y[0], oo), p[np++] = point(x, y[1], oo);
            p[np++] = point(x, y[2], oo), p[np++] = point(x, y[3], oo);
            w[nw++] = Wall(point(x,   -1), point(x, y[0]));
            w[nw++] = Wall(point(x, y[1]), point(x, y[2]));
            w[nw++] = Wall(point(x, y[3]), point(x,   11));
        }
        p[np++] = point(10, 5, oo);

        for(i=1; i<np; i++)
        for(j=0; j<i;  j++)
        {
            point t = p[i] - p[j];
            t.len = sqrt(t.x*t.x+t.y*t.y);

            if(p[i].len > t.len + p[j].len && CanLine(p[i], p[j], w, nw) == true)
                p[i].len = t.len + p[j].len;
        }

        printf("%.2f\n", p[np-1].len);
    }

    return 0;
}