1、对每条边松弛|V|-1次
2、解决单源最短路径问题
3、一般情况，变得权重可以为负
4、时间复杂度O（V*E）
伪码：
BELLMAN-FORD(G,w,S)
INITIALIZE-SINGLE-SOURCR(G,S) 1、初始化所有节点
for i=1 to |G.V|-1 2、进行循环
for each edge(u,v)属于G.E
RELAX(u,v,w) 对每条边都进行一次松弛
for each edge(u,v)属于G.E 3、判断是否存在权值为负的回路
if v.d > u.d+w(u,v)
return FALSE
return TRUE
C++实现
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 << "输入节点的个数:"; cin >> nodenum;
cout << "输入边的条数:"; cin >> edgenum;
cout << "输入源节点的编号:"; cin >> source;
for (int i = 1; i <= nodenum; ++i)
dist[i] = MAXINT;
dist[source] = 0;
cout << "输入" << 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){
//判断是否有负权重值的回路
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》
本文出自 “零蛋蛋” 博客，谢绝转载！
原文地址：http://lingdandan.blog.51cto.com/10697032/1924752
