题意

给你n个点，1为起点，2为终点，要求所有1到2所有路径中每条路径上最大值的最小值。

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思路

不想打最短路
跑一边最小生成树，再扫一遍1到2的路径，取最大值即可

注意g++要用%f输出！！！

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常数巨大的丑陋代码

# include <stdio.h>

# include <stdlib.h>

# include <iostream>

# include <string.h>

# include <math.h>

# include <algorithm>

using namespace std;

# define IL inline

# define RG register

# define UN unsigned

# define ll long long

# define rep(i, a, b) for(RG int i = a; i <= b; i++)

# define per(i, a, b) for(RG int i = b; i >= a; i--)

# define uev(e, u) for(RG int e = ft[u]; e != -1; e = edge[e].nt)

# define mem(a, b) memset(a, b, sizeof(a))

# define max(a, b) (((a) > (b)) ? (a) : (b))

# define min(a, b) (((a) < (b)) ? (a) : (b))

# define Swap(a, b) a ^= b, b ^= a, a ^= b;

# define Sqr(a) ((a) * (a))

IL ll Get(){

    RG char c = '!'; RG ll x = 0, z = 1;

    while(c != '-' && (c < '0' || c > '9')) c = getchar();

    if(c == '-') z = -1, c = getchar();

    while(c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();

    return x*z;

}

const int MAXN = 1000001;

int ft[MAXN], n, cnt, vis[MAXN], fa[MAXN], m;

struct Edge{

    int to, nt;

    double f;

} edge[MAXN<<1];

struct _Edge{

    int u, v;

    double f;

    IL bool operator < (_Edge b) const{

        return f < b.f;

    }

} _edge[MAXN];

double ans, x[MAXN], y[MAXN];

IL void Add(RG int u, RG int v, RG double f){

    edge[cnt] = (Edge){v, ft[u], f}; ft[u] = cnt++;

}

IL int Find(RG int x){

    return fa[x] == x ? x : Find(fa[x]);

}

IL void Kruskal(){

    sort(_edge + 1, _edge + m + 1);

    RG int t = 0;

    rep(i, 1, m){

        if(t == n - 1) break;

        RG int u = Find(_edge[i].u), v = Find(_edge[i].v);

        if(u != v){

            t++; fa[u] = v;

            Add(u, v, _edge[i].f); Add(v, u, _edge[i].f);

        }

    }

}

IL void Dfs(RG int u, RG double ans1){

    if(u == 2) ans = ans1;

    uev(e, u){

        RG int v = edge[e].to;

        if(vis[v]) continue;

        vis[v] = 1;

        RG double f = max(ans1, edge[e].f);

        Dfs(v, f);

    }

}

int main(){

    RG int T = 0;

    n = Get();

    while(n){

        T++;

        mem(ft, -1); m = ans = cnt = 0;

        rep(i, 1, n) scanf("%lf%lf", &x[i], &y[i]);

        rep(i, 1, n-1) rep(j, i+1, n){

            double dis = sqrt(Sqr(x[i] - x[j]) + Sqr(y[i] - y[j]));

            _edge[++m] = (_Edge){i, j, dis};

        }

        rep(i, 1, n) fa[i] = i;

        Kruskal();

        mem(vis, 0); vis[1] = 1;

        Dfs(1, 0);

        printf("Scenario #%d\nFrog Distance = %.3f\n\n", T, ans);

        n = Get();

    }

    return 0;

}