洛谷P4779 【模板】单源最短路径(标准版) 题解 Dijkstra+堆优化

题目链接:https://www.luogu.com.cn/problem/P4779

参考博客:https://www.cnblogs.com/-Wind-/p/10164910.html

这里用的是优先队列,时间复杂度 \(O(m \cdot \text{log }m)\) ,实现代码如下:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010, maxm = 200020;
const int INF = (1<<29);
struct Edge {
    int v, w, nxt;
    Edge() {};
    Edge(int _v, int _w, int _nxt) { v = _v; w = _w; nxt = _nxt; }
} edge[maxm];
int n, m, s, head[maxn], ecnt;
void init() {
    ecnt = 0;
    memset(head, -1, sizeof(int)*(n+1));
}
void addedge(int u, int v, int w) {
    edge[ecnt] = Edge(v, w, head[u]); head[u] = ecnt ++;
}
struct Node {
    int u, dis;
    Node() {};
    Node(int _u, int _dis) { u = _u; dis = _dis; }
    bool operator < (const Node x) const {
        return dis > x.dis;
    }
};
priority_queue<Node> que;
int dis[maxn];
bool vis[maxn];
void dijkstra_pq() {
    memset(dis, -1, sizeof(int)*(n+1));
    dis[s] = 0;
    que.push(Node(s, 0));
    while (!que.empty()) {
        Node nd = que.top();
        que.pop();
        int u = nd.u;
        if (vis[u]) continue;
        vis[u] = true;
        for (int i = head[u]; i != -1; i = edge[i].nxt) {
            int v = edge[i].v, w = edge[i].w;
            if (dis[v] == -1 || dis[v] > nd.dis + w) {
                dis[v] = nd.dis + w;
                que.push(Node(v, dis[v]));
            }
        }
    }
}
int main() {
    scanf("%d%d%d", &n, &m, &s);
    init();
    while (m --) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        addedge(u, v, w);
    }
    dijkstra_pq();
    for (int i = 1; i <= n; i ++) printf("%d ", dis[i]);
    return 0;
}

应该算最暴力的 “Dijkstra+堆优化”。

关于“Dijkstra+堆优化”的时间复杂度

看用什么堆,手写二叉堆是O(elogv),stl优先队列是O(eloge),斐波那契堆是O(vlogv+e),配对堆复杂度玄学

参考链接:https://tieba.baidu.com/p/5547681010?red_tag=1480858706

posted @ 2020-05-21 01:26  quanjun  阅读(160)  评论(0编辑  收藏  举报