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;
}