POJ 1106 Transmitters （简单计算几何）

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博客原文地址：http://blog.csdn.net/xuechelingxiao/article/details/39297079

Transmitters

题目大意：给你一个半圆的圆心跟半径，再给你N个点，半圆可以绕圆心旋转任意角度，求半圆最多可以覆盖的点的个数是多少。

解题思路：因为圆心是固定的，就很简单了。先把在圆的覆盖范围内的点找出来，再对这些点循环去找对于每个点来说，跟它在同一侧的点的个数，同侧的点判断就用叉积就可以，当叉积>=0的时候就是同一侧的。

代码如下：

复制代码

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#define sqr(x) ((x)*(x))
using namespace std;
const double eps = 1e-8;

int dcmp(double x) {
    return x < -eps ? -1 : x > eps;
}

struct Point {
    double x, y;
} P[155];

double Distance(Point a, Point b){
    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}

double xmult(Point a, Point b, Point c) {
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}

int main()
{
    Point o;
    double R;
    int n;
    while(~scanf("%lf%lf%lf", &o.x, &o.y, &R)) {
        if(dcmp(R) < 0) {
            break;
        }
        scanf("%d", &n);
        Point p;
        int m = 0;
        for(int i = 0; i < n; ++i) {
            scanf("%lf%lf", &p.x, &p.y);
            if(dcmp(Distance(p, o)-R) <= 0) {
                P[m++] = p;
            }
        }
        int ans = 0;
        for(int i = 0; i < m; ++i) {
            int num = 0;
            for(int j = 0; j < m; ++j) {
                if(dcmp(xmult(P[i], P[j], o)) >= 0) {
                    num++;
                }
            }
            //printf("p%d = %lf %lf num = %dn", i, P[i].x, P[i].y, num);
            ans = max(ans, num);
        }
        printf("%dn", ans);
    }

return 0;
}

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#include <stdio.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#define sqr(x) ((x)*(x))
using namespace std;
const double eps = 1e-8;

int dcmp(double x) {
    return x < -eps ? -1 : x > eps;
}

struct Point {
    double x, y;
} P[155];

double Distance(Point a, Point b){
    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}

double xmult(Point a, Point b, Point c) {
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}

int main()
{
    Point o;
    double R;
    int n;
    while(~scanf("%lf%lf%lf", &o.x, &o.y, &R)) {
        if(dcmp(R) < 0) {
            break;
        }
        scanf("%d", &n);
        Point p;
        int m = 0;
        for(int i = 0; i < n; ++i) {
            scanf("%lf%lf", &p.x, &p.y);
            if(dcmp(Distance(p, o)-R) <= 0) {
                P[m++] = p;
            }
        }
        int ans = 0;
        for(int i = 0; i < m; ++i) {
            int num = 0;
            for(int j = 0; j < m; ++j) {
                if(dcmp(xmult(P[i], P[j], o)) >= 0) {
                    num++;
                }
            }
            //printf("p%d = %lf %lf num = %dn", i, P[i].x, P[i].y, num);
            ans = max(ans, num);
        }
        printf("%dn", ans);
    }

    return 0;
}