Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists‘ sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona‘s stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog‘s jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy‘s stone, Fiona‘s stone and all other stones in the lake. Your job is to compute the frog distance between Freddy‘s and Fiona‘s stone.
Input
The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy‘s stone, stone #2 is Fiona‘s stone, the other n-2 stones are unoccupied. There‘s a blank line following each test case. Input is terminated by a value of zero (0) for n.
Output
For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.
Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

題意 ： 要求青蛙從第一個點跳到第二個點，其他的點作為跳板，可以選擇跳或者不跳。問最小權值中的最大值是多少。

註意 ： 這題wA 了好久，因為精度的問題，c++交可以過，g++就wa了

　　　還有 sqrt(x), 裏面的 x得是浮點型的數，不然交的時候給你提示編譯錯誤

代碼：

const int inf = 1<<29;
struct qnode
{
    double x, y;
}arr[205];
int n;
double fun(int i, int j){
    double ff = sqrt((arr[i].x - arr[j].x)*(arr[i].x - arr[j].x) + (arr[i].y - arr[j].y)*(arr[i].y - arr[j].y));
    return ff;
}
double edge[205][205];
double ans;

struct node
{
    int v;
    double c;
    node(int _v, double _c):v(_v), c(_c){}
    friend bool operator< (node n1, node n2){
        return n1.c > n2.c;
    }
};
double d[205];
bool vis[205];

void prim(){
    ans = 0;
    priority_queue<node>que;
    while(!que.empty()) que.pop();
    for(int i = 1; i <= n; i++){
        d[i] = edge[1][i];
        que.push(node(i, d[i]));
    }
    memset(vis, false, sizeof(vis));    
    while(!que.empty()){
        node tem = que.top();
        que.pop();
        int v = tem.v;
        double c = tem.c;
        
        ans = max(ans, c);
        if (v == 2) return;
        if (vis[v]) continue;
        vis[v] = true;;
        for(int i = 1; i <= n; i++){
            if (!vis[i] && edge[v][i] < d[i]){
                d[i] = edge[v][i];
                que.push(node(i, d[i]));
            }
        }
    }
}

int main() {

    int k = 1;
    while(~scanf("%d", &n) && n){
        for(int i = 1; i <= n; i++){
            scanf("%lf%lf", &arr[i].x, &arr[i].y);
        }   
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++)
                edge[i][j] = inf;
        }
        for(int i = 1; i <= n; i++){
            for(int j = i+1; j <= n; j++){
                double len = fun(i, j);
                edge[i][j] = edge[j][i] = len; 
            }
        }
        prim();
        printf("Scenario #%d\n", k++);
        printf("Frog Distance = %.3f\n\n", ans);
    }
    return 0;
}

poj 2253 最短路 or 最小生成樹