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11018: Mars Buggy Problem

11018: Mars Buggy Problem. ★★★☆☆ 題組： Problem Set Archive with Online Judge 題號： 11018: Mars Buggy Problem 解題者： 鍾正一 解題日期： 200 7 年 4 月 17 日 題意： 給定 N 個殖民地的經度以及緯度，接者輸入 M 組起點、終點以及每次車子可以跑的最遠距離， 每到一個殖民地，車子的油量就會捕充滿 ， 利用程式判斷是否存在最短路徑，若有列出經過的路線，若無，列出最接近的距離。. Output: Scenario 1:

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11018: Mars Buggy Problem

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  1. 11018: Mars Buggy Problem • ★★★☆☆ • 題組：Problem Set Archive with Online Judge • 題號： 11018: Mars Buggy Problem • 解題者：鍾正一 • 解題日期：2007年4月17日 • 題意：給定N個殖民地的經度以及緯度，接者輸入M組起點、終點以及每次車子可以跑的最遠距離，每到一個殖民地，車子的油量就會捕充滿，利用程式判斷是否存在最短路徑，若有列出經過的路線，若無，列出最接近的距離。

  2. Output: Scenario 1: ------------------------------ From Lousberg to van_den_Hoogen with range 1200 km: Lousberg at 0 km. van_de_Kieft at 1198 km. Lubbers at 2154 km. Duponselle at 3065km. van_den_Hoogen at 3969 km. ------------------------------ From Rasschaert to Ramnath with range 1000 km: No route for this range, minimum required range is 1217 km. ------------------------------ From Lubbers to van_Dijk with range 10 km: Lubbers at 0 km. van_Dijk at 5 km. ------------------------------ • 題意範例： Input: 11 Lousberg 0.500000 1.000000 Rasschaert 0.000000 0.500000 Lubbers 0.000000 1.000000 van_den_Hoogen -0.500000 1.000000 Bink 0.000000 1.500000 van_de_Kieft 0.200000 0.800000 Bronkhorst -0.300000 1.100000 van_Dijk 0.001000 1.001000 Zijlstra 0.010000 1.020000 Duponselle -0.250000 0.900000 Ramnath -0.400000 0.600000 3 Lousberg van_den_Hoogen 1200 Rasschaert Ramnath 1000 Lubbers van_Dijk 1000

  3. 解法：利用題目所給的距離計算公式 • a = haversine(lat2 - lat1) • b = cos(lat1) * cos(lat2) * haversine(lon2 - lon1) • c = 2 * atan2(sqrt(a + b), sqrt(1 - a - b)) • d = R * c • 將所有點的距離都算出來，並且把超過車子行走最大距離的路徑刪除，再找出最短的距離。 • 解法範例：無 • 討論： (1) 範圍:2<=n<=100,1<=m<=100,地名不超過128個字母 (2) 緯度+-π/2 ，經度0~2π (3)所有的距離計算後都四捨五入 (4) 時間複雜度O(n2)

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