poj2253 Frogger

题目：

Description
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
空行
3
17 4
19 4
18 5
空行
0
Sample Output
Scenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414

题意：
一只青蛙想从自己脚下的石头开始调到另一只青蛙的石头上，它可以直接跳，也可以借助别的空石头跳。输入的数据是每个石头的坐标（第一个是这只青蛙踩的石头的坐标，第二个是另一只青蛙踩的石头的坐标）。
问最短的青蛙距离（从一块石头到另一块石头的最大距离）。
实际上就是求最短路径中的前后两个点的距离中的最大值。
实际上这道题就是求最小生成树中的最大权值。
由于点较少，可以用Floyd-Warshall算法（求任意两点间的最短距离）。

参考代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int MAXV = 200+10;
const int MAXE = 200000+10;
const double INF = 99999999;

struct node {
    int x, y, count;
} n[MAXV];

double diss[MAXV][MAXV];

void init(int n1) {
    memset(n, 0, sizeof(n));
    for (int i = 1;i <= n1;++i) {
        for (int j = 1;j <= n1;++j) {
            diss[i][j] = INF;
        }
    }
    for (int i = 1;i <= n1;++i) {
        diss[i][i] = 0;
    }
}

double disjuli(node a1, node a2) {
    double ans = sqrt((double)((a1.x - a2.x) * (a1.x - a2.x)) + (double)((a1.y - a2.y) * (a1.y - a2.y)));
    return ans;
}

double floyd(int n1) {
    for (int k = 1;k <= n1;++k) {
        for (int i = 1;i <= n1;++i) {
            for (int j = 1;j <= n1;++j) {
                diss[i][j] = min(diss[i][j], max(diss[i][k], diss[k][j]));
            }
        }
    }
    return diss[1][2];
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n1;
    int cnt = 1;
    while (cin >> n1) {
        if (n1 == 0) break;
        init(n1);       
        for (int i = 1;i <= n1;++i) {
            cin >> n[i].x >> n[i].y;
            n[i].count = i;
        }
        for (int i = 1;i <= n1;++i) {
            for (int j = 1;j <= n1;++j) {
                if (i != j) {
                    double disjuli1 = disjuli(n[i], n[j]);
                    diss[n[i].count][n[j].count] = disjuli1;
                }
            }
        }
        double ans = floyd(n1);
        cout << "Scenario #" << cnt << endl;
        cout << "Frog Distance = " << fixed << setprecision(3) << ans << endl;
        cout << endl;
        cnt++;
    }
    return 0;
}