最短路变形。。。。

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

坑爹的英文题。。  这里要求的是到达Fiona所有path的最小必须跳跃距离（the largest edge of paths） 
关键分析 我们用d[v]表示 起点到v点的所有path的最小必须跳跃距离
那么就有状态转移方程 d[v]=for(j=1~n) if(d[j]<d[v]&&cost(j->v)<d[v]) d[v]=max(cost,d[j])

状态方程出来， 其实就是最短路的改版。，。 实现过程几乎没有差别、
上代码：
kee point: 1.tey to comprehend the meanings of text!
           2. design the stata transition.
 #include<algorithm>
    #include<queue>
    #include<stdio.h>
    #include<string.h>
    #include<vector>
    #include<math.h>
    using namespace std;

    const int maxn = 205;
    const int oo = 0xfffffff;

    struct point{double x, y;}p[maxn];
    struct node
    {
        int y;
        double len;
        node(int y, double len):y(y), len(len){}
    };
    vector<node> G[maxn];
    double v[maxn];

    //求两点间的距离
    double Len(point a, point b)
    {
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    }
    void Spfa()
    {
        queue<int> Q;
        Q.push(1);

        while(Q.size())
        {
            int s = Q.front();Q.pop();
            int len = G[s].size();

            for(int i=0; i<len; i++)
            {
                node q = G[s][i];

                if(v[s] < v[q.y] && q.len < v[q.y])
                {
                    v[q.y] = max(v[s], q.len);//要选取一条路上的最大的那条边
                    Q.push(q.y);
                }
            }
        }
    }

    int main()
    {
        int N, t=1;

        while(scanf("%d", &N), N)
        {
            int i, j;

            for(i=1; i<=N; i++)
                scanf("%lf%lf", &p[i].x, &p[i].y);

            for(i=1; i<=N; i++)
            for(j=1; j<=N; j++)
            {
                if(i == j)
                    continue;

                double len = Len(p[i], p[j]);
                G[i].push_back(node(j, len));
            }

            for(i=1; i<=N; i++)
                v[i] = oo;
            v[1] = 0;

            Spfa();

            if(t != 1)printf("\n");
            printf("Scenario #%d\n", t++);
            printf("Frog Distance = %.3f\n", v[2]);

            for(i=1; i<=N; i++)
                G[i].clear();
        }

        return 0;