LCP 35

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。

示例 1：

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]

输出：43

解释：最佳路线为：1->3->0。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。充电用时共 310 + 51= 35 行驶用时 3 + 5 = 8，此时总用时最短 43。

示例 2：

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]

输出：38

解释：最佳路线为：0->4->3->2。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。充电用时 42+28+3*1 = 27 行驶用时 2+5+4 = 11，总用时最短 38。

提示：

1 <= paths.length <= 200 paths[i].length == 3 2 <= charge.length == n <= 100 0 <= path[i][0],path[i][1],start,end < n 1 <= cnt <= 100 1 <= path[i][2] <= cnt 1 <= charge[i] <= 100 题目保证所有城市相互可以到达

Solutions

dijkstra

class Solution {
public:
#define node(c, pos) ((pos + 1) * 101 + (c + 1))        
    int electricCarPlan(vector<vector<int>>& paths, int cnt, int start, int end, vector<int>& charge) {
        using tii = tuple<long, int, int>;

        vector<vector<pair<int, int>>> g(100);
        for (auto & e : paths) {
            int st = e[0], ed = e[1], dis = e[2];
            g[st].emplace_back(ed, dis);
            g[ed].emplace_back(st, dis);
        }

        vector<bool> visited(101 * 120, false);
        vector<long> dis(101 * 120, 0x3f3f3f3f);
        priority_queue<tii, vector<tii>, greater<>> pq;
        dis[node(0, start)] = 0;
        pq.emplace(0, 0, start);

        while (!pq.empty()) {
            auto [time, c, cur] = pq.top(); pq.pop();
            if (visited[node(c, cur)]) continue;
            if (c == 0 && cur == end) break;
            visited[node(c, cur)] = true;
            // two options
            // 1. stay and charge
            for (int nc = c + 1; nc <= cnt; nc++) {
                int new_time = time + (nc - c) * charge[cur];
                if (new_time < dis[node(nc, cur)]) {
                    dis[node(nc, cur)] = new_time;
                    pq.emplace(new_time, nc, cur);
                }
            }
            // 2. go to the next place
            for (auto & [out, d] : g[cur]) {
                if (c < d) continue;
                int next = node(c - d, out);
                int new_time = time + d;
                if (new_time < dis[next]) {
                    dis[next] = new_time;
                    pq.emplace(new_time, c - d, out);
                }
            }

        }

        return dis[node(0, end)];
    }
};

PreviousLCP 17 NextLCP 34

Last updated 4 years ago

Was this helpful?

Solutions

dijkstra

class Solution {
public:
#define node(c, pos) ((pos + 1) * 101 + (c + 1))        
    int electricCarPlan(vector<vector<int>>& paths, int cnt, int start, int end, vector<int>& charge) {
        using tii = tuple<long, int, int>;

        vector<vector<pair<int, int>>> g(100);
        for (auto & e : paths) {
            int st = e[0], ed = e[1], dis = e[2];
            g[st].emplace_back(ed, dis);
            g[ed].emplace_back(st, dis);
        }

        vector<bool> visited(101 * 120, false);
        vector<long> dis(101 * 120, 0x3f3f3f3f);
        priority_queue<tii, vector<tii>, greater<>> pq;
        dis[node(0, start)] = 0;
        pq.emplace(0, 0, start);

        while (!pq.empty()) {
            auto [time, c, cur] = pq.top(); pq.pop();
            if (visited[node(c, cur)]) continue;
            if (c == 0 && cur == end) break;
            visited[node(c, cur)] = true;
            // two options
            // 1. stay and charge
            for (int nc = c + 1; nc <= cnt; nc++) {
                int new_time = time + (nc - c) * charge[cur];
                if (new_time < dis[node(nc, cur)]) {
                    dis[node(nc, cur)] = new_time;
                    pq.emplace(new_time, nc, cur);
                }
            }
            // 2. go to the next place
            for (auto & [out, d] : g[cur]) {
                if (c < d) continue;
                int next = node(c - d, out);
                int new_time = time + d;
                if (new_time < dis[next]) {
                    dis[next] = new_time;
                    pq.emplace(new_time, c - d, out);
                }
            }

        }

        return dis[node(0, end)];
    }
};