CSU 1412 Line and Circles

　　原题链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1412

　　题目要求判断是否有一条直线可以穿过所有的圆。

　　做法：把所有圆心做一次凸包，然后判断这个凸包是否能通过一个宽度为2*R的通道。

　　做法和求凸包直径差不多，只是判断的时候把点到两个端点的距离换成点到直线的距离。

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

#include <algorithm>

using namespace std;

#define inf 1e20

#define eps 1e-8
const int N = 100005 ;

const double PI = 2.0*asin(1.0); //高精度求PI

struct Lpoint

{

    double x,y;

}a[N], b[N]; //点

struct Llineseg

{

    Lpoint a,b;

}; //线段

struct Ldir

{

    double dx,dy;

}; //方向向量

struct Lline

{

    Lpoint p;

    Ldir dir;

}; //直线
bool mult(Lpoint sp, Lpoint ep, Lpoint op)

{

    return (sp.x - op.x) * (ep.y - op.y)

           >= (ep.x - op.x) * (sp.y - op.y);

}
bool operator < (const Lpoint &l, const Lpoint &r)

{

    return l.y < r.y || (l.y == r.y && l.x < r.x);

}
int graham(Lpoint pnt[], int n, Lpoint res[])

{

    int i, len, top = 1;

    sort(pnt, pnt + n);

    if (n == 0) return 0;

    res[0] = pnt[0];

    if (n == 1) return 1;

    res[1] = pnt[1];

    if (n == 2) return 2;

    res[2] = pnt[2];

    for (i = 2; i < n; i++)

    {

        while (top && mult(pnt[i], res[top], res[top-1]))

            top--;

        res[++top] = pnt[i];

    }

    len = top;

    res[++top] = pnt[n - 2];

    for (i = n - 3; i >= 0; i--)

    {

        while (top!=len && mult(pnt[i], res[top], res[top-1])) top--;

        res[++top] = pnt[i];

    }

    return top; // 返回凸包中点的个数

}
void format(Lline ln, double& A, double& B, double& C)

{

    A=ln.dir.dy;

    B=-ln.dir.dx;

    C=ln.p.y*ln.dir.dx-ln.p.x*ln.dir.dy;
}
double p2ldis(Lpoint a, Lline ln)

{

    double A,B,C;

    format(ln,A,B,C);

    return(fabs(A*a.x+B*a.y+C)/sqrt(A*A+B*B));

}
double CPMD(Lpoint p[], int n)//ConvexPolygonMinimumDiameter

{

    int i, j;

    double ans = inf, tmp;

    p[n] = p[0];

    Lline ln;

    Ldir dir;

    for(i = 0, j = 1; i < n; ++i)

    {

        if((i+1)%n == j) j = (j + 1) % n;

        dir.dx = p[i].x - p[i+1].x;

        dir.dy = p[i].y - p[i+1].y;

        ln.dir = dir;

        ln.p = p[i];

        while((tmp = p2ldis(p[j], ln)) < (p2ldis(p[(j+1)%n], ln)))

            j = (j + 1) % n;

        ans = min(ans, tmp);

    }

    return ans;

}
double dis(Lpoint u, Lpoint v)

{

    return sqrt((u.x-v.x) * (u.x-v.x) + (u.y - v.y)*(u.y - v.y));

}
int main()

{

    int n, t;

    double r;

    scanf("%d", &t);

    while(t--)

    {

        scanf("%d%lf", &n, &r);

        for(int i = 0; i < n; i++)

            scanf("%lf%lf", &a[i].x, &a[i].y);

        int m = graham(a, n, b);

        if(m <= 2)

        {

            printf("Yes\n");

            continue;

        }

        double k = CPMD(b, m);

        if(k - 2*r < eps)

            printf("Yes\n");

        else

            printf("No\n");

    }

    return 0;

}

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