POJ 3227 Mountains 两线段交点模板

题意：两人登高望远，所能看到的山的长度。

设两人所处位置为v，则有对于p[i]，若其在射线vp[i-1]上方，则线段p[i]p[i-1]都能看见，若在下方或线上，则记录下当起点site = i-1；

然后判断后面的点p[i]与射线vp[site]的位置关系。

PS：貌似以前写的线段相交是错的，不知为啥还能A掉题。。。。。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#define LL long long
#define PI (acos(-1))
#define EPS (1e-4)

using namespace std;

struct P
{
    double x,y;
} p[1100];

double X_Mul(P a1,P a2,P b1,P b2)
{
    P v1 = {a2.x-a1.x,a2.y-a1.y},v2 = {b2.x-b1.x,b2.y-b1.y};
    return v1.x*v2.y - v1.y*v2.x;
}

double Cal_Point_Dis(P p1,P p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
}

P Cal_Segment_Cross_Point(P a1,P a2,P b1,P b2)
{
    double t = X_Mul(b1,b2,b1,a1) / X_Mul(a1,a2,b1,b2);
    P cp = {a1.x+(a2.x-a1.x)*t,a1.y+(a2.y-a1.y)*t};
    return cp;
}

double Cal_Visible_Surface(P *p,P v,int n)
{
    int i = 1,site;
    double len = 0;

    while(i < n)
    {
        for(; i < n && X_Mul(v,p[i],v,p[i-1]) < -EPS; ++i)
        {
            len += Cal_Point_Dis(p[i],p[i-1]);
        }

        for(site = i-1; i < n; ++i)
        {
            if(X_Mul(v,p[site],v,p[i]) > EPS)
            {
                len += Cal_Point_Dis(p[i],Cal_Segment_Cross_Point(v,p[site],p[i],p[i-1]));
                site = i;
                if((i&1) == 0)
                {
                    ++i;
                    break;
                }
            }
        }
    }
    return len;
}

int main()
{
    int n,i;
    P v = {0,0};
    while(scanf("%d %lf",&n,&v.y) && (v.y || n))
    {
        for(i = 0; i < n; ++i)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
        }

        printf("%.2lf\n",Cal_Visible_Surface(p,v,n));

    }
    return 0;
}