B

Think:
1知识点：spfa()算法求最短路
2题意：n = 1 的青蛙要跳到 n = 2 的青蛙哪里，可以其它石头作为中介，询问在可达路径中青蛙至少要能跳多远

建议参考博客1
建议参考博客2

vjudge题目链接

以下为Accepted代码

#include <cstdio>#include <cstring>#include <queue>#include <cmath>using namespace std;const int N = 240;const int inf = 0x3f3f3f3f;int n, x[N], y[N], vis[N];double e[N][N], dis[N];void spfa();int main(){    int k = 1;    while(scanf("%d", &n) && n){        memset(e, 0, sizeof(e));        for(int i = 1; i <= n; i++)            scanf("%d %d", &x[i], &y[i]);        for(int i = 1; i <= n; i++){            for(int j = i; j <= n; j++){                double x_2 = (double)((x[i]-x[j])*(x[i]-x[j]));                double y_2 = (double)((y[i]-y[j])*(y[i]-y[j]));                e[i][j] = e[j][i] = sqrt(x_2 + y_2);            }        }        spfa();        printf("Scenario #%d\n", k++);        printf("Frog Distance = %.3lf\n", dis[2]);        printf("\n");    }    return 0;}void spfa(){    queue <int> q;    for(int i = 1; i <= n; i++)        dis[i] = inf;    ///memset(dis, inf, sizeof(dis));不能将double类型的数组初始化为无穷大    memset(vis, 0, sizeof(vis));    q.push(1);    dis[1] = 0, vis[1] = 1;    while(!q.empty()){        int t1 = q.front();        q.pop();        vis[t1] = 0;        for(int v = 1; v <= n; v++){            if(max(dis[t1], e[t1][v]) < dis[v]){                dis[v] = max(dis[t1], e[t1][v]);                if(vis[v] == 0){                    q.push(v);                    vis[v] = 1;                }            }        }    }}