02253: Frogger

http://cs101.openjudge.cn/dsapre/02253/

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.

输入

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.

输出

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.

样例输入

样例输出

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

来源

Ulm Local 1997

python

"""
定义了一个frog_distance函数，它接受一个石头列表，并计算出青蛙从第一个石头跳到第二个石头的最小跳跃距离。
使用动态规划的思想，通过计算任意两个石头之间的直线距离，并利用最短路径算法（Floyd-Warshall算法）计算出最小跳跃距离。

在主循环中，读取输入并计算每个测试用例的青蛙距离。当输入的石头数量为0时，循环终止。

输出每个测试用例的结果，包括测试用例的序号和青蛙距离，保留3位小数。在每个测试用例后输出一个空行。
"""
import math

def frog_distance(stones):
    n = len(stones)
    distances = [[float('inf')] * n for _ in range(n)]

    for i in range(n):
        for j in range(n):
            if i == j:
                distances[i][j] = 0
            else:
                x1, y1 = stones[i]
                x2, y2 = stones[j]
                distance = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)
                distances[i][j] = distance

    for k in range(n):
        for i in range(n):
            for j in range(n):
                distances[i][j] = min(distances[i][j], max(distances[i][k], distances[k][j]))

    return distances[0][1]

# 读取输入
test_case = 1
while True:
    n = int(input())
    if n == 0:
        break

    stones = []
    for _ in range(n):
        x, y = map(int, input().split())
        stones.append((x, y))

    # 计算青蛙距离
    distance = frog_distance(stones)

    # 输出结果
    print("Scenario #{}".format(test_case))
    print("Frog Distance = {:.3f}".format(distance))
    print()
    input()

    test_case += 1

02253: Frogger ​

02253: Frogger