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POJ 3525 Most Distant Point from the Sea 半平面交 +二分

2014年09月05日 ⁄ 综合 ⁄ 共 1106字 ⁄ 字号 评论关闭

 

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#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define eps 1e-8
#define inf 1<<29

struct point
{
    double x, y;
}p[105], tp[105], pp[105];

double a, b, c;
int n, m;

void getline(point p1, point p2)
{
    a = p2.y - p1.y;
    b = p1.x - p2.x;
    c = p2.x * p1.y - p2.y * p1.x;
}

point intersect(point p1, point p2)
{
    double u = fabs(a * p1.x + b * p1.y + c);
    double v = fabs(a * p2.x + b * p2.y + c);
    point ret;
    ret.x = (v * p1.x + u * p2.x) / (u + v);
    ret.y = (v * p1.y + u * p2.y) / (u + v);
    return ret;
}

void cut()
{
    int i, tm = 0;
    for(i = 1; i <= m; i++)
    {
        if( a * pp[i].x + b * pp[i].y + c <= 0)
            tp[++tm] = pp[i];
        else 
        {
            if(a * pp[i-1].x + b * pp[i-1].y + c < 0)
                tp[++tm] = intersect(pp[i-1], pp[i]);
            if(a * pp[i+1].x + b * pp[i+1].y + c < 0)
                tp[++tm] = intersect(pp[i], pp[i+1]);
        }
    }
    for(i = 1; i <= tm; i++)
        pp[i] = tp[i];
    pp[0] = pp[tm];
    pp[tm+1] = p[1];
    m = tm;
}

bool solve(double mid)
{
    int i;
    for(i = 1; i <= n; i++)
        pp[i] = p[i];
    pp[0] = pp[n];
    pp[n+1] = pp[1];
    p[n+1] = p[1];
    m = n;
    for(i = 1; i <= n; i++)
    {
        getline(p[i], p[i+1]);
        c += mid * sqrt(a * a + b * b); 
        cut();
        if(m == 0) break;
    }
    return m > 0;
}

int main()
{
    int i;
    while( ~scanf("%d", &n) && n)
    {
        for(i = 1; i <= n; i++)
            scanf("%lf%lf", &p[i].x, & p[i].y);
        double l = 0, r = inf;
        while(r - l >= 1e-6)
        {
            double mid = (l + r)/2;
            if( solve(mid) ) l = mid;
            else r = mid;
        }
        printf("%.6f\n", l);
    }
    return 0;
}

 

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