2018ACM-ICPC南京現場賽D題-Country Meow

Problem D. Country Meow
Input file: standard input
Output file: standard output
In the 24th century, there is a country somewhere in the universe, namely Country Meow. Due to advanced technology, people can easily travel in the 3-dimensional space.
There are N cities in Country Meow. The i-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 combatant command, so that troops in different cities can easily communicate. Hence, the Euclidean distance between the 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 integer N (1 ≤ N ≤ 100).
The following N lines describe the i-th city located.Each line contains three integers xi, yi, zi(−100000 ≤ xi, yi, zi ≤ 100000).
Output
Print a real number — the minimum Euclidean distance between the combatant command and the farthest city. Your answer is considered correct if its absolute or relative error does not exceed 10−3. Formally, let your answer be a, and the jury’s answer be b. Your answer is considered correct if |a−b| max(1,|b|) ≤ 10−3.
standard input
3
0 0 0
3 0 0
0 4 0
4
0 0 0
1 0 0
0 1 0
0 0 1
standard output
2.500000590252103
0.816496631812619

思路：

題意是最小球覆蓋，一定要讀懂題。
好像是計算幾何板子題，不過三個三分也是可以過的，模擬退火玄學演算法不清楚。

AC_CODE：

#include <bits/stdc++.h>
#define o2(x) (x)*(x)
using namespace std;
typedef long long LL;
const int MXN = 1e5 + 5;
int n;
int x[MXN], y[MXN], z[MXN];
double len(double X, double Y, double Z, int i) {
    return o2(X-x[i])+o2(Y-y[i])+o2(Z-z[i]);
}
double exe3(double X, double Y, double Z) {
    double ans = 0;
    for(int i = 1; i <= n; ++i) ans = max(ans, len(X,Y,Z,i));
    return ans;
}
double exe2(double X, double Y) {
    double l = -1e6, r = 1e6, midl, midr, ans;
    for(int i = 0; i < 70; ++i) {
        midl = (l+r)/2;
        midr = (midl+r)/2;
        if(exe3(X, Y, midl) <= exe3(X, Y, midr)) {
            r = midr, ans = midl;
        }else {
            l = midl, ans = midr;
        }
    }
    return exe3(X, Y, ans);
}
double exe1(double X) {
    double l = -1e6, r = 1e6, midl, midr, ans;
    for(int i = 0; i < 70; ++i) {
        midl = (l+r)/2;
        midr = (midl+r)/2;
        if(exe2(X, midl) <= exe2(X, midr)) {
            r = midr, ans = midl;
        }else {
            l = midl, ans = midr;
        }
    }
    return exe2(X, ans);
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d%d%d", &x[i], &y[i], &z[i]);
    double l = -1e6, r = 1e6, midl, midr, ans;
    for(int i = 0; i < 70; ++i) {
        midl = (l+r)/2;
        midr = (midl+r)/2;
        if(exe1(midl) <= exe1(midr)) {
            r = midr, ans = midl;
        }else {
            l = midl, ans = midr;
        }
    }
    double tmp = exe1(ans);
    printf("%.9f\n", sqrt(tmp));
    return 0;
}