pku 2253(floyd变形)

/*
  Name: floyd 
  Copyright: 
  Author: 
  Date: 19/04/12 22:39
  Description: 求所有连接1跟2的路径中，最大边权中的最小值 
*/

#include <cmath>
#include <cstdio>
#include <iostream>

using namespace std;

const int N = 205;

double dis[N][N];

struct point {
    double x;
    double y;
}p[N];

double max(double a, double b) {
    return a > b ? a : b;
}

double floyd(int n) {
    for (int k=0; k<n; ++k) {
        for (int i=0; i<n; ++i) {
            for (int j=0; j<n; ++j) {
                double maxs = max(dis[i][k], dis[k][j]);//递推式变形 
                if (dis[i][j] > maxs) dis[i][j] = maxs;
            }
        }
    }
    return dis[0][1];
}

int main() {
    int n, t = 0;
    while (scanf("%d", &n), n) {
        for (int i=0; i<n; ++i) scanf ("%lf%lf", &p[i].x, &p[i].y);
        for (int i=0; i<n; ++i) {
            for (int j=0; j<i; ++j) {
                if (i == j) dis[i][j] = 0;
                else dis[i][j] = dis[j][i] = sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y)); 
            }
        }
        double ans = floyd(n);
        printf ("Scenario #%d\n", ++t);
        printf ("Frog Distance = %.3lf\n\n", ans);
    }
    return 0;
}