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"question": {
"questionId": "1000220",
"questionFrontendId": "LCP 35",
"categoryTitle": "Algorithms",
"boundTopicId": 713325,
"title": "电动车游城市",
"titleSlug": "DFPeFJ",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "电动车游城市",
"translatedContent": "小明的电动车电量充满时可行驶距离为 `cnt`,每行驶 1 单位距离消耗 1 单位电量,且花费 1 单位时间。小明想选择电动车作为代步工具。地图上共有 N 个景点,景点编号为 0 ~ N-1。他将地图信息以 `[城市 A 编号,城市 B 编号,两城市间距离]` 格式整理在在二维数组 `paths`,表示城市 A、B 间存在双向通路。初始状态,电动车电量为 0。每个城市都设有充电桩`charge[i]` 表示第 i 个城市每充 1 单位电量需要花费的单位时间。请返回小明最少需要花费多少单位时间从起点城市 `start` 抵达终点城市 `end`。\n\n\n**示例 1**\n>输入:`paths = [[1,3,3],[3,2,1],[2,1,3],[0,1,4],[3,0,5]], cnt = 6, start = 1, end = 0, charge = [2,10,4,1]`\n>\n>输出:`43`\n>\n>解释最佳路线为1->3->0。\n>在城市 1 仅充 3 单位电至城市 3然后在城市 3 充 5 单位电,行驶至城市 5。\n>充电用时共 3\\*10 + 5\\*1= 35\n>行驶用时 3 + 5 = 8此时总用时最短 43。\n![image.png](https://pic.leetcode-cn.com/1616125304-mzVxIV-image.png)\n\n\n\n\n**示例 2**\n>输入:`paths = [[0,4,2],[4,3,5],[3,0,5],[0,1,5],[3,2,4],[1,2,8]], cnt = 8, start = 0, end = 2, charge = [4,1,1,3,2]`\n>\n>输出:`38`\n>\n>解释最佳路线为0->4->3->2。\n>城市 0 充电 2 单位,行驶至城市 4 充电 8 单位,行驶至城市 3 充电 1 单位,最终行驶至城市 2。\n>充电用时 4\\*2+2\\*8+3\\*1 = 27\n>行驶用时 2+5+4 = 11总用时最短 38。\n\n**提示:**\n- `1 <= paths.length <= 200`\n- `paths[i].length == 3`\n- `2 <= charge.length == n <= 100`\n- `0 <= path[i][0],path[i][1],start,end < n`\n- `1 <= cnt <= 100`\n- `1 <= path[i][2] <= cnt`\n- `1 <= charge[i] <= 100`\n- 题目保证所有城市相互可以到达",
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"lang": "C++",
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"code": "class Solution {\npublic:\n int electricCarPlan(vector<vector<int>>& paths, int cnt, int start, int end, vector<int>& charge) {\n\n }\n};",
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"code": "class Solution(object):\n def electricCarPlan(self, paths, cnt, start, end, charge):\n \"\"\"\n :type paths: List[List[int]]\n :type cnt: int\n :type start: int\n :type end: int\n :type charge: List[int]\n :rtype: int\n \"\"\"",
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