文件目录

#include <bits/stdc++.h>
using namespace std;
double n, x[200], y[200], g[200][200], maxl[200];
char c;
double getl(int i, int j)
{
    return sqrt(pow(x[i] - x[j], 2) + pow(y[i] - y[j], 2));
}
int main()
{
    cin >> n;
    // 邻接矩阵存图
    for (int i = 1; i <= n; i++)
        cin >> x[i] >> y[i];
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
        {
            cin >> c;
            if (c == '1')
                g[i][j] = getl(i, j);
            else
                g[i][j] = 1e10;
        }
    // floyd求最短路
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (i != j && i != k && j != k && g[i][k] < 1e10 && g[k][j] < 1e10 && g[i][j] > g[i][k] + g[k][j])
                {
                    g[i][j] = g[i][k] + g[k][j];
                }
    // maxl保存所有点在图中的直径
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (g[i][j] != 1e10)
                maxl[i] = max(maxl[i], g[i][j]);
    // 假设ij连通，找最小新直径
    double ans = 2e10;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (i != j && g[i][j] == 1e10)
            {
                double t = getl(i, j);
                ans = min(ans, maxl[i] + maxl[j] + t);
            }
    for (int i = 1; i <= n; i++)
        ans = max(maxl[i], ans); // 存疑，同学理解了分享给老师
    printf("%.6lf", ans);
    return 0;
}