poj2074Line of Sight(直线相交）

链接

几何细节题。

对于每一个障碍物可以求出它在地产线上的覆盖区间，如下图。

紫色部分即为每个障碍物所覆盖掉的区间，求出所有的，扫描一遍即可。

几个需要注意的地方：直线可能与地产线没有交点，可视区间可能包含地产线的端点，扫描的时候保留当前扫到的最大值。

代码中的数据很经典，供参考。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 10100
#define LL long long
#define INF 0xfffffff
#define zero(x) (((x)>0?(x):-(x))<eps)
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y) {}
} p[N];
struct line
{
    point u,v;
} li[N];
typedef point pointt;
pointt operator -(point a,point b)
{
    return point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
bool cmp(point a,point b)
{
    if(dcmp(a.x-b.x)==0)
        return a.y<b.y;
    return a.x<b.x;
}

bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交
{
    double a1, b1, c1, a2, b2, c2, d;
    a1 = p1.y - p2.y;
    b1 = p2.x - p1.x;
    c1 = p1.x*p2.y - p2.x*p1.y;
    a2 = p3.y - p4.y;
    b2 = p4.x - p3.x;
    c2 = p3.x*p4.y - p4.x*p3.y;
    d = a1*b2 - a2*b1;
    if (!dcmp(d))    return false;
    p.x = (-c1*b2 + c2*b1) / d;
    p.y = (-a1*c2 + a2*c1) / d;
    return true;
}
int main()
{
    int i,n;
    int x1,x2,y;
    while(scanf("%d%d%d",&x1,&x2,&y)&&x1&&x2&&y)
    {
        li[1].u = point(x1,y);
        li[1].v = point(x2,y);
        li[1].v.y = li[1].u.y;
        scanf("%lf%lf%lf",&li[2].u.x,&li[2].v.x,&li[2].u.y);
        li[2].v.y = li[2].u.y;
        scanf("%d",&n);
        for(i = 3; i <= n+2; i++)
        {
            scanf("%lf%lf%lf",&li[i].u.x,&li[i].v.x,&li[i].u.y);
            li[i].v.y = li[i].u.y;
        }
        n+=2;
        int g = 0;
        for(i = 3 ; i <= n; i++)
        {
            if(dcmp(li[i].u.y-li[1].u.y)>=0||dcmp(li[i].u.y-li[2].u.y)<=0) continue;
            point pp;
            intersection1(li[1].u,li[i].v,li[2].u,li[2].v,pp);
            p[++g].y = min(li[2].v.x,max(pp.x,li[2].u.x));
            intersection1(li[1].v,li[i].u,li[2].u,li[2].v,pp);
            p[g].x = max(li[2].u.x,min(li[2].v.x,pp.x));
        }
        sort(p+1,p+g+1,cmp);
        double maxz,ty ;
       // cout<<p[1].x<<" "<<p[1].y<<endl;
        if(g==0)
        maxz = li[2].v.x-li[2].u.x;
        else
        {
            maxz = max(p[1].x-li[2].u.x,li[2].v.x-p[g].y);
            ty = p[1].y;
        }

        for(i = 1; i < g; i++)
        {
           // printf("%.3f %.3f\n",p[i].x,p[i].y);
            ty = max(ty,p[i].y);
            if(p[i+1].x>ty)
            {
                maxz = max(p[i+1].x-ty,maxz);//printf("%.3f %.3f\n",ty,maxz);
                ty = p[i+1].y;
            }

        }
        if(dcmp(maxz)<=0)
            puts("No View");
        else
            printf("%.2f\n",maxz);
    }
    return 0;
}
/*
2 6 6
0 15 0
3
1 2 1
3 4 1
12 13 1
1 5 5
0 10 0
1
0 15 1
2 6 6
0 15 0
3
1 2 1
3 4 1
12 13 1
2 6 6
0 15 0
4
1 2 1
3 4 1
12 13 1
1 5 2
2 6 6
0 15 0
2
0 5 3
6 15 3
2 6 6
0 15 0
2
6 10 1
0 2 1
2 6 6
0 15 0
1
2 6 7
2 6 6
0 15 0
1
2 6 7
2 6 6
0 15 0
1
4 4.5 5.5
2 6 6
0 15 0
16
0 1 3
1.5 2 3
2.5 3 3
3.5 4 3
4.5 5 3
5.5 6 3
6.5 7 3
7.5 8 3
8.5 9 3
9.5 10 3
10.5 11 3
11.5 12 3
12.5 13 3
13.5 14 3
14.5 15 3
15.5 16 3
2 6 6
0 15 0
16
0 1 .1
1.5 2 .1
2.5 3 .1
3.5 4 .1
4.5 5 .1
5.5 6 .1
6.5 7 .1
7.5 8 .1
8.5 9 .1
9.5 10 .1
10.5 11 .1
11.5 12 .1
12.5 13 .1
13.5 14 .1
14.5 15 .1
15.5 16 .1
2 6 6
0 15 0
14
0 1 3
1.5 2 3
2.5 3 3
3.5 4 3
4.5 5 3
5.5 6 3
8.5 9 3
9.5 10 3
10.5 11 3
11.5 12 3
12.5 13 3
13.5 14 3
14.5 15 3
15.5 16 3
2 6 6
0 4000000000 0
2
1 2 1 
15 16 3
2 6 6
0 15 1
5
1 1.5 6
17 18 1 
3 5 3
0 20 10
0 20 0.5

答案：
8.80
No View
8.80
6.70
No View
4.00
15.00
15.00
No View
No View
0.44
1.00
3999999970.00
8.00
*/