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 = ;
const double oo = 1e9+;
const double EPS = 1e-;

struct point
{
    double x, y, len;
    point(double x=, double y=, double len=):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 ;
        if(fabs(c)<EPS)return ;
        return -;
    }
};
struct Wall
{
    point A, B;
    Wall(point A=, point B=):A(A), B(B){}
};

bool CanLine(point a, point b, Wall w[], int N)
{
    for(int i=; 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 == )
            return false;
    }

    return true;
}

int main()
{
    int M;

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

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

            p[np++] = point(x, y[], oo), p[np++] = point(x, y[], oo);
            p[np++] = point(x, y[], oo), p[np++] = point(x, y[], oo);
            w[nw++] = Wall(point(x,   -), point(x, y[]));
            w[nw++] = Wall(point(x, y[]), point(x, y[]));
            w[nw++] = Wall(point(x, y[]), point(x,   ));
        }
        p[np++] = point(, , oo);

        for(i=; i<np; i++)
        for(j=; 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-].len);
    }

    return ;
}