最小球覆盖&最小圆覆盖 题目总结
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最小圆覆盖：
HYSBZ 1337 最小圆覆盖
给出平面上N个点,N<=10^5.请求出一个半径最小的圆覆盖住所有的点
Input
第一行给出数字N，现在N行，每行两个实数x,y表示其坐标.
Output
输出最小半径，输出保留三位小数.
Sample Input
4 1 0 0 1 0 -1 -1 0
Sample Output
1.000
AC代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long lt;
#define eps 1e-6
#define sqr(x) ((x)*(x))
const int maxn=1000010;
int n;
struct point{
double x,y;
}p[maxn],O;
double R;//半径
double getd(point a,point b){ //求直径
return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}
point getO(point p1,point p2,point p3) { //求圆心
point res;
double a=p2.x-p1.x;
double b=p2.y-p1.y;
double c=p3.x-p2.x;
double d=p3.y-p2.y;
double e=sqr(p2.x)+sqr(p2.y)-sqr(p1.x)-sqr(p1.y);
double f=sqr(p3.x)+sqr(p3.y)-sqr(p2.x)-sqr(p2.y);
res.x=(f*b-e*d)/(c*b-a*d)/2.0;
res.y=(a*f-e*c)/(a*d-b*c)/2.0;
return res;
}
void mincir() {
O=p[1];
R=0;
for(int i=1;i<=n;++i){
if(getd(p[i],O)-R>eps) { //不在圆内
O=p[i];
R=0;
for(int j=1;jeps) {//不在圆内
O=(point){(p[i].x+p[j].x)/2.0,(p[i].y+p[j].y)/2.0};
R=getd(p[i],p[j])/2.0;
for(int k=1;keps) {//不在圆内
O=getO(p[i],p[j],p[k]); //外接圆
R=getd(p[i],O);
}
}
}
}
}
}
int main()
{
cin>>n;
for(int i=1;i<=n;++i)
scanf("%lf%lf",&p[i].x,&p[i].y);
random_shuffle(p+1,p+1+n);// random_shuffle()随机打乱函数 首指针 尾指针
mincir();
printf("%.3f",R);
return 0;
}
剩下几题都套这个板子改一下输入输出稳过
HDU 3007 Buried memory
HYSBZ 1336 Alien最小圆覆盖
HYSBZ 2823 信号塔
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最小球覆盖：
POJ 2069 Super Star
During a voyage of the starship Hakodate-maru (see Problem 1406), researchers found strange synchronized movements of stars. Having heard these observations, Dr. Extreme proposed a theory of "super stars". Do not take this term as a description of actors or singers. It is a revolutionary theory in astronomy.
According to this theory, starts we are observing are not independent objects, but only small portions of larger objects called super stars. A super star is filled with invisible (or transparent) material, and only a number of points inside or on its surface shine. These points are observed as stars by us.
In order to verify this theory, Dr. Extreme wants to build motion equations of super stars and to compare the solutions of these equations with observed movements of stars. As the first step, he assumes that a super star is sphere-shaped, and has the smallest possible radius such that the sphere contains all given stars in or on it. This assumption makes it possible to estimate the volume of a super star, and thus its mass (the density of the invisible material is known).
You are asked to help Dr. Extreme by writing a program which, given the locations of a number of stars, finds the smallest sphere containing all of them in or on it. In this computation, you should ignore the sizes of stars. In other words, a star should be regarded as a point. You may assume the universe is a Euclidean space.
Input
The input consists of multiple data sets. Each data set is given in the following format.
n
x1 y1 z1
x2 y2 z2
. . .
xn yn zn
The first line of a data set contains an integer n, which is the number of points. It satisfies the condition 4 <= n <= 30.
The location of n points are given by three-dimensional orthogonal coordinates: (xi, yi, zi) (i = 1, ..., n). Three coordinates of a point appear in a line, separated by a space character. Each value is given by a decimal fraction, and is between 0.0 and 100.0 (both ends inclusive). Points are at least 0.01 distant from each other.
The end of the input is indicated by a line containing a zero.
Output
For each data set, the radius of the smallest sphere containing all given points should be printed, each in a separate line. The printed values should have 5 digits after the decimal point. They may not have an error greater than 0.00001.
Sample Input
4 10.00000 10.00000 10.00000 20.00000 10.00000 10.00000 20.00000 20.00000 10.00000 10.00000 20.00000 10.00000 4 10.00000 10.00000 10.00000 10.00000 50.00000 50.00000 50.00000 10.00000 50.00000 50.00000 50.00000 10.00000 0
Sample Output
7.07107 34.64102
#include
#include
#include
#include
#include
using namespace std;
const double eps=1e-5;
struct POINT{
double x,y,z;
}p[110],op;//N个点
int n;
double dist(POINT &a,POINT &b){//两点距离
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
double solve(){
double ret,delta=100.0;//温度
double maxDis,tempDis;
int i,id;
while(delta>eps){
id=0;
maxDis=dist(op,p[id]);
for(i=1;imaxDis){
maxDis=tempDis;
id=i;
}
}
ret=maxDis;
op.x+=(p[id].x-op.x)/maxDis*delta;
op.y+=(p[id].y-op.y)/maxDis*delta;
op.z+=(p[id].z-op.z)/maxDis*delta;
delta*=0.98;
}
return ret;//最小球半径
}
int main(){
while(scanf("%d",&n)!=EOF&&n){
for(int i=0;i
Problem D. Country Meow
In the 24th century, there is a country somewhere in the universe, namely Country Meow. Due to advancedtechnology, people can easily travel in the 3-dimensional space.There areNcities in Country Meow. Thei-th city is located at(xi, yi, zi)in Cartesian coordinate.Due to the increasing threat from Country Woof, the president decided to build a new combatantcommand, so that troops in different cities can easily communicate. Hence, the Euclidean distance betweenthe combatant command and any city should be minimized.Your task is to calculate the minimum Euclidean distance between the combatant command and the farthest city.
Input
The first line contains an integerN(1≤N≤100).The followingNlines describe thei-th city located.Each line contains three integersxi, yi, zi(−100000≤xi, yi, zi≤100000).
Output
Print a real number — the minimum Euclidean distance between the combatant command and the farthestcity. Your answer is considered correct if its absolute or relative error does not exceed10−3. Formally, letyour answer bea, and the jury’s answer beb. Your answer is considered correct if|a−b|max(1,|b|)≤10−3.Examplesstandard inputstandard output30 0 03 0 00 4 02.50000059025210340 0 01 0 00 1 00 0 10.816496631812619
#include
#include
#include
#include
using namespace std;
const double eps=1e-8;
struct point3D
{
double x,y,z;
} data[105];
int n;
double dis(point3D a,point3D b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
double solve()
{
double step=10000,ans=1e30,mt;
point3D z;
z.x=z.y=z.z=0;
int s=0;
while(step>eps)
{
for(int i=0; i