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圖論-Dijkstra堆優化

之前已經寫過樸素的Dijkstra了:

int Dijkstra(int x) { 
    int min, k;  
    for(int i=1; i<=n; i++) d[i] = a[x][i];  
    d[x] = 0;  
    InS[x] = true;  
    for(int i=1; i<=n-1; i++) {
        min = 1e9;  
        for(int j=1; j<=n; j++) 
            if(min > d[j] && !InS[j])
                min = d[k=j];  
        InS[k] = true;  
        for(int j=1; j<=n; j++)
            if(d[j] > d[k]+a[k][j])  
                d[j] = d[k] + a[k][j];  
    }  
    for(int i=1; i<=n; i++) cout << d[i] << " ";   
}  

下面是基於vector陣列的堆優化Dijkstra

#include <iostream>
#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;

struct Edge {
	int from, to, dis;
	Edge(int u, int v, int w) : from(u), to(v), dis(w) {}
};

struct HeapNode {
	int d, u;
	bool operator <(const HeapNode h) const {
		return d > h.d;
	}
};

#define MAXN 10010

struct Dijkstra {
	int n, m;
	vector<Edge> edges; //所有邊
	vector<int> G[MAXN]; //以i為起點的所有邊
	bool done[MAXN];
	int d[MAXN];
	int p[MAXN]; //記錄上一條弧
	void init(int n) {
		this -> n = n;
		for(int i=1; i<=n; i++) G[i].clear();
		edges.clear();
	}
	void Add(int u, int v, int d) {
		edges.push_back(Edge(u, v, d));
		m = edges.size();
		G[u].push_back(m-1);
	}
	void dijkstra(int s) {
		priority_queue<HeapNode> Q;
		for(int i=1; i<=n; i++) d[i] = 2147483647;
		d[s] = 0;
		memset(done, 0, sizeof(done));
		Q.push((HeapNode) {
			0, s
		});
		while(!Q.empty()) {
			HeapNode x = Q.top();
			Q.pop();
			int u = x.u;
			if(done[u]) continue;
			done[u] = true;
			for(int i=0; i<G[u].size(); i++) {
				Edge e = edges[G[u][i]];
				if(d[e.to] > d[u] + e.dis) {
					d[e.to] = d[u] + e.dis;
					p[e.to] = G[u][i]; //記錄最短路徑
					Q.push((HeapNode) {
						d[e.to], e.to
					});
				}
			}
		}
	}
	void Print_Path(int x) { //源點到x的最短路徑
		if(x == 0) return;
		Print_Path(p[x]);
		printf("%d ->", x);
	}
} Dijk;

int main() {
	int n, m, u, v, w, s;
	scanf("%d%d%d", &n, &m, &s);
	Dijk.init(n);
	for(int i=1; i<=m; i++) {
		scanf("%d%d%d", &u, &v, &w);
		Dijk.Add(u, v, w);
	}
	Dijk.dijkstra(s);
	for(int i=1; i<=n; i++) printf("%d ", Dijk.d[i]);
	return 0;
}