1 条题解
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0
改造道路(Revamping Trails G) 题解
思路分析
按题目要求使用对应数据结构与算法模拟/递推求解,注意数据范围边界与时间复杂度。
参考代码
#include <iostream> #include <vector> #include <queue> #include <tuple> #include <algorithm> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, m, k; if (!(cin >> n >> m >> k)) return 0; vector<vector<pair<int, long long>>> adj(n + 1); for (int i = 0; i < m; ++i) { int u, v; long long w; cin >> u >> v >> w; adj[u].push_back({v, w}); adj[v].push_back({u, w}); } const long long INF = (1LL << 58); vector<vector<long long>> dist(n + 1, vector<long long>(k + 1, INF)); using State = tuple<long long, int, int>; priority_queue<State, vector<State>, greater<State>> pq; dist[1][0] = 0; pq.push({0, 1, 0}); while (!pq.empty()) { auto [d, u, j] = pq.top(); pq.pop(); if (d != dist[u][j]) continue; for (auto& edge : adj[u]) { int v = edge.first; long long w = edge.second; if (d + w < dist[v][j]) { dist[v][j] = d + w; pq.push({dist[v][j], v, j}); } if (j < k && d < dist[v][j + 1]) { dist[v][j + 1] = d; pq.push({d, v, j + 1}); } } } long long ans = INF; for (int j = 0; j <= k; ++j) ans = min(ans, dist[n][j]); cout << ans << "\n"; return 0; }
- 1
信息
- ID
- 8792
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者