题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1775
详细题解:很好的一道计算几何题目,考的就是想法,想到了就很简单了……
先算出每两个圆心间的距离,dis[i][j]表示第i和第j个圆心的距离
然后再以每条线段ij为边,求出以这样的线段为直线的左右两边的点的个数cnt和距离len[cnt],直线上的点要同时属于两边。 然后就可以排下len数组的序,ans = min(ans, len[m]);就ok了; 还有就是注意一下ans可能会小于0的情况。
G++不能用%lf真是个深坑……
#include<cstdio> #include<cmath> #include<climits> #include<iostream> #include<algorithm> using namespace std; #define N 210 #define eps 1e-10 double dis[N][N]; struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) {} }; Point p[N]; typedef Point Vector; Vector operator - (Point A, Point B) {return Vector(A.x-B.x, A.y-B.y);} double Dot(Vector A, Vector B){return A.x*B.x + A.y*B.y;} double Length(Vector A){return sqrt(Dot(A, A));} double Cross(Vector A, Vector B){return A.x*B.y - A.y*B.x; } int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } int main () { int n, m; while(scanf("%d %d", &n, &m) != EOF) { for(int i = 1; i <= n; i++) scanf("%lf %lf", &p[i].x, &p[i].y); if(m <= 2) { printf("0.000000\n"); continue;} for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) dis[i][j] = Length(p[i] - p[j]); double ans = INT_MAX; for(int i = 1; i <= n; i++) for(int j = i+1; j <= n; j++) { double len_l[N], len_r[N]; int cnt_l = 0, cnt_r = 0; for(int k = 1; k <= n; k++) { double tmp = Cross(p[j]-p[i], p[k]-p[i]); int ha = dcmp(tmp); if(ha == 0) { cnt_l++;cnt_r++; len_l[cnt_l] = len_r[cnt_r] = 0; } else if(ha > 0) { cnt_l++; len_l[cnt_l] = fabs(tmp) / dis[i][j]; } else{ cnt_r++; len_r[cnt_r] = fabs(tmp)/ dis[i][j]; } } sort(len_l+1, len_l+cnt_l+1); sort(len_r+1, len_r+cnt_r+1); if(cnt_l >= m) ans = min(ans, len_l[m]); if(cnt_r >= m) ans = min(ans, len_r[m]); } ans--; if(ans <= 0) ans = 0; printf("%.10f\n", ans); } return 0; }