mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
162 lines
19 KiB
JSON
162 lines
19 KiB
JSON
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"questionId": "1000437",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1425673,
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"title": "夺回据点",
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"translatedTitle": "夺回据点",
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"translatedContent": "欢迎各位勇者来到力扣城，本次试炼主题为「夺回据点」。\n\n魔物了占领若干据点，这些据点被若干条道路相连接，`roads[i] = [x, y]` 表示编号 `x`、`y` 的两个据点通过一条道路连接。\n\n现在勇者要将按照以下原则将这些据点逐一夺回：\n\n- 在开始的时候，勇者可以花费资源先夺回一些据点，初始夺回第 `j` 个据点所需消耗的资源数量为 `cost[j]` \n\n- 接下来，勇者在不消耗资源情况下，每次可以夺回**一个**和「已夺回据点」相连接的魔物据点，并对其进行夺回\n\n> 注：为了防止魔物暴动，勇者在每一次夺回据点后（包括花费资源夺回据点后），需要保证剩余的所有魔物据点之间是相连通的（不经过「已夺回据点」）。\n\n请返回勇者夺回所有据点需要消耗的最少资源数量。\n\n**注意：**\n- 输入保证初始所有据点都是连通的，且不存在重边和自环\n\n**示例 1：**\n>输入：\n>`cost = [1,2,3,4,5,6]`\n>`roads = [[0,1],[0,2],[1,3],[2,3],[1,2],[2,4],[2,5]]`\n>\n>输出：`6`\n>\n>解释：\n>勇者消耗资源 `6` 夺回据点 `0` 和 `4`，魔物据点 `1、2、3、5` 相连通；\n>第一次夺回据点 `1`，魔物据点 `2、3、5` 相连通；\n>第二次夺回据点 `3`，魔物据点 `2、5` 相连通；\n>第三次夺回据点 `2`，剩余魔物据点 `5`；\n>第四次夺回据点 `5`，无剩余魔物据点；\n>因此最少需要消耗资源为 `6`，可占领所有据点。\n{:height=170px}\n\n\n**示例 2：**\n>输入：\n>`cost = [3,2,1,4]`\n>`roads = [[0,2],[2,3],[3,1]]`\n>\n>输出：`2`\n>\n>解释：\n>勇者消耗资源 `2` 夺回据点 `1`，魔物据点 `0、2、3` 相连通；\n>第一次夺回据点 `3`，魔物据点 `2、0` 相连通；\n>第二次夺回据点 `2`，剩余魔物据点 `0`；\n>第三次夺回据点 `0`，无剩余魔物据点；\n>因此最少需要消耗资源为 `2`，可占领所有据点。\n{:height=60px}\n\n\n**提示：**\n- `1 <= roads.length, cost.length <= 10^5`\n- `0 <= roads[i][0], roads[i][1] < cost.length`\n- `1 <= cost[i] <= 10^9`\n",
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