Bellman-Ford&SPFA算法
<Bellman-Ford算法>
Dijkstra算法中要求边的权非负，如果遇到负权，则可以采用Bellman-Ford算法。
适用条件&范围
1.单源最短路径(从源点s到其它所有顶点v);
2.有向图&无向图(无向图可以看作(u,v),(v,u)同属于边集E的有向图);
3.边权可正可负(如有负权回路输出错误提示);
4.差分约束系统;
算法描述
1.对每条边进行|V|-1次Relax操作; ---------- |V| 表示节点的个数
2.如果存在(u,v)∈E使得dis[u]+w<dis[v],则存在负权回路;否则dis[v]即为s到v的最短距离,pre [v]为前驱。-----------dis[u] 表示经过第 1 步后s到u的最短距离,w表示某边权值,pre[v]所表示的就是上一次更新dis[v]的中间点即第 1 步中的u点
伪代码
-------------------PASCAL------------
For i:=1 to |V|-1 do
For 每条边(u,v)∈E do
Relax(u,v,w);
For每条边(u,v)∈E do
If dis[u]+w<dis[v] Then Exit(False)
-----------------C&C++--------------
void bellman_ford(int v)
{
for 1 to n
initialize dist[v]; //此时只经过一条边
for 2 to n-1 (i) //经过的边数不大于i条时
for 1 to n (j) //对于每一个目标顶点
for 1 to n (k) //经过该顶点时，与当前最小值比较，并更新当前值
if edge[k][j] > 0 && dist[k] > edge[k][j]+dist[j]
更新当前值
}
时空复杂度
算法时间复杂度O(VE)。因为算法简单，适用范围又广，虽然复杂度稍高，仍不失为一个很实用的算法。
参考代码
----------------PASCAL-----------------
{单源最短路径的Bellman-ford算法
执行v-1次，每次对每条边进行松弛操作
如有负权回路则输出"Error"
}
const
maxn=100;
maxe=maxn*(maxn-1)div 2;
type
edge=record
a,b,w :integer;
end;
var
edges :array[1..maxe]of edge;
dis :array[1..maxn]of integer;
pre :array[1..maxn]of integer;
e,n,s :integer;
procedure init;
var
i :integer;
begin
e:=0;
assign(input,'g.in');reset(input);
readln(n,s);
while not eof do
begin
inc(e);
with edges[e] do readln(a,b,w);
end;
fillchar(dis,sizeof(dis),$7f);//$7f是什么，解释替换 $7f 是127 $在pascal中代表后面的数是16进制
dis[s]:=0;pre[s]:=s;
end;
procedure relax(u,v,w:integer);
begin
if dis[u]+w<dis[v] then
begin
dis[v]:=dis[u]+w;
pre[v]:=u;
end
end;
function bellman_ford:boolean;
var
i,j :integer;
begin
for i:=1 to n-1 do
for j:=1 to e do
with edges[j] do relax(a,b,w);
for i:=1 to e do
with edges[i] do
if dis[a]+w<dis[b] then exit(false);
exit(true)
end;
procedure print_path(i:integer);
begin
if pre[i]<>s then print_path(pre[i]);
write('-->',i)
end;
procedure show;
var
i :integer;
begin
for i:=1 to n do
begin
write(i:3,':',dis[i]:3,':',s);
print_path(i);
writeln
end;
end;
{========main========}
begin
init;
if bellman_ford then show
else writeln('Error!!')
end.
--------------------Matlab-------------
function ford(d,n,s) % d为已知图的邻接矩阵，n为顶点数（各顶点标号为1，2...,n），s为源点标号
for i=1:n %初始化dist，pre
dist(i)=inf; %dist（i）为s，i之间的最短路的长度
pre(i)=NaN; %pre（i）为s到i的最短路上i的前一个顶点
end
dist(s)=0;
for k=1:n-1
for i=1:n %松弛操作
for j=1:n
if d(i,j)~=inf
if dist(j)>dist(i)+d(i,j)
dist(j)=dist(i)+d(i,j);
pre(j)=i;
end
end
end
end
end
for i=1:n
for j=1:n
if d(i,j)~=inf
if dist(i)+d(i,j)<dist(j)%判断有无负权回路
error('negetive weight circut');
end
end
end
end
dist
pre
end
引申：SPFA算法
算法简介
SPFA(Shortest Path Faster Algorithm)是Bellman-Ford算法的一种队列实现，减少了不必要的冗余计算。
算法流程
算法大致流程是用一个队列来进行维护。 初始时将源加入队列。 每次从队列中取出一个元素，并对所有与他相邻的点进行松弛，若某个相邻的点松弛成功，则将其入队。 直到队列为空时算法结束。
算法代码
Procedure SPFA;
Begin
initialize-single-source(G,s);
initialize-queue(Q);
enqueue(Q,s);
while not empty(Q) do begin
u:=dequeue(Q);
for each v∈adj[u] do begin
tmp:=d[v];
relax(u,v);
if (tmp<>d[v]) and (not v in Q) then enqueue(v);
end;
end;
End;
POJ3259
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 7131 | Accepted: 2499 |
Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output
Sample Input
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8
Sample Output
NO YES
Hint
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
Source