一、對每條邊鬆弛|V|-1次node
二、解決單源最短路徑問題ide
三、通常狀況,變得權重能夠爲負spa
四、時間複雜度O(V*E)ci
僞碼:it
BELLMAN-FORD(G,w,S)class
INITIALIZE-SINGLE-SOURCR(G,S) 一、初始化全部節點循環
for i=1 to |G.V|-1 二、進行循環di
for each edge(u,v)屬於G.E 時間
RELAX(u,v,w) 對每條邊都進行一次鬆弛view
for each edge(u,v)屬於G.E 三、判斷是否存在權值爲負的迴路
if v.d > u.d+w(u,v)
return FALSE
return TRUE
C++實現
typedef struct Edge{ int src; int dst; int weight; }Edge; int nodenum; int edgenum; int source; const int MAXINT = 9999; const int MAXNUM = 100; Edge edge[MAXNUM]; int dist[MAXNUM]; void init() { cout << "輸入節點的個數:"; cin >> nodenum; cout << "輸入邊的條數:"; cin >> edgenum; cout << "輸入源節點的編號:"; cin >> source; for (int i = 1; i <= nodenum; ++i) dist[i] = MAXINT; dist[source] = 0; cout << "輸入" << edgenum << "行src dst weight"; for (int i = 1; i <= edgenum; ++i){ cin >> edge[i].src >> edge[i].dst >> edge[i].weight; if (edge[i].src == source){ dist[edge[i].dst] = edge[i].weight; } } } bool bellmanFord() { init(); for (int i = 1; i <= nodenum - 1; ++i){ for (int j = 1; j <= edgenum; ++j){ //relax int u = edge[j].src; int v = edge[j].dst; int weight = edge[j].weight; if (dist[v] > dist[u] + weight) dist[v] = dist[u] + weight; } } for (int i = 1; i <= edgenum; ++i){ //判斷是否有負權重值的迴路 int u = edge[i].src; int v = edge[i].dst; int weight = edge[i].weight; if (dist[v] > dist[u] + weight) return false; } return true; } int main() { if (bellmanFord()){ for (int i = 1; i <= nodenum; ++i) cout << dist[i] << " "; cout << endl; } system("pause"); return 0; }
《完》