單源最短路徑,非負權重
單源最短路徑演算法(對於非負權重)也稱為 Dijkstra 演算法。給定一個圖 G(V,E) 及其鄰接矩陣表示,並提供一個源頂點。Dijkstra 演算法用於查詢從源頂點到圖 G 中任何其他頂點的最小最短路徑。
從起始節點到任何其他節點,找到最小的距離。在這個問題中,圖使用鄰接矩陣表示。(對於此目的,成本矩陣和鄰接矩陣是相似的)。
輸入 - 鄰接矩陣 -
0 3 6 ∞ ∞ ∞ ∞ 3 0 2 1 ∞ ∞ ∞ 6 2 0 1 4 2 ∞ ∞ 1 1 0 2 ∞ 4 ∞ ∞ 4 2 0 2 1 ∞ ∞ 2 ∞ 2 0 1 ∞ ∞ ∞ 4 1 1 0
輸出 -
0 to 1, Using: 0, Cost: 3 0 to 2, Using: 1, Cost: 5 0 to 3, Using: 1, Cost: 4 0 to 4, Using: 3, Cost: 6 0 to 5, Using: 2, Cost: 7 0 to 6, Using: 4, Cost: 7
演算法
dijkstraShortestPath(n, dist, next, start)
輸入 - 節點總數 n,每個頂點的距離列表,儲存下一個節點的 next 列表,以及種子或起始頂點。
輸出 - 從起始節點到所有其他頂點的最短路徑。
Begin create a status list to hold the current status of the selected node for all vertices u in V do status[u] := unconsidered dist[u] := distance from source using cost matrix next[u] := start done status[start] := considered, dist[start] := 0 and next[start] := φ while take unconsidered vertex u as distance is minimum do status[u] := considered for all vertex v in V do if status[v] = unconsidered then if dist[v] > dist[u] + cost[u,v] then dist[v] := dist[u] + cost[u,v] next[v] := u done done End
示例(C++)
#include<iostream> #define V 7 #define INF 999 using namespace std; //Cost matrix of the graph int costMat[V][V] = { {0, 3, 6, INF, INF, INF, INF}, {3, 0, 2, 1, INF, INF, INF}, {6, 2, 0, 1, 4, 2, INF}, {INF, 1, 1, 0, 2, INF, 4}, {INF, INF, 4, 2, 0, 2, 1}, {INF, INF, 2, INF, 2, 0, 1}, {INF, INF, INF, 4, 1, 1, 0} }; int minimum(int *status, int *dis, int n){ int i, min, index; min = INF; for(i = 0; i<n; i++) if(dis[i] < min && status[i] == 1){ min = dis[i]; index = i; } if(status[index] == 1) return index;//minimum unconsidered vertex distance else return -1;//when all vertices considered } void dijkstra(int n, int *dist,int *next, int s){ int status[V]; int u, v; //initialization for(u = 0; u<n; u++){ status[u] = 1;//unconsidered vertex dist[u] = costMat[u][s];//distance from source next[u] = s; } //for source vertex status[s] = 2; dist[s] = 0; next[s] = -1;//-1 for starting vertex while((u = minimum(status, dist, n)) > -1){ status[u] = 2;//now considered for(v = 0; v<n; v++) if(status[v] == 1) if(dist[v] > dist[u] + costMat[u][v]){ dist[v] = dist[u] + costMat[u][v];//update distance next[v] = u; } } } main(){ int dis[V], next[V], i, start = 0; dijkstra(V, dis, next, start); for(i = 0; i<V; i++) if(i != start) cout << start << " to " << i <<", Using: " << next[i] << ", Cost: " << dis[i] << endl; }
輸出
0 to 1, Using: 0, Cost: 3 0 to 2, Using: 1, Cost: 5 0 to 3, Using: 1, Cost: 4 0 to 4, Using: 3, Cost: 6 0 to 5, Using: 2, Cost: 7 0 to 6, Using: 4, Cost: 7
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