1、對每條邊鬆弛|V|-1次
2、解決單源最短路徑問題
3、一般情況，變得權重可以爲負
4、時間複雜度O（V*E）
僞碼：
BELLMAN-FORD(G,w,S)
INITIALIZE-SINGLE-SOURCR(G,S) 1、初始化所有節點
for i=1 to |G.V|-1 2、進行循環
for each edge(u,v)屬於G.E
RELAX(u,v,w) 對每條邊都進行一次鬆弛
for each edge(u,v)屬於G.E 3、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》