最短路径
A.标号法求解单源点最短路径:
var
a:array[1..maxn,1..maxn] of integer;
b:array[1..maxn] of integer; {b指顶点i到源点的最短路径}
mark:array[1..maxn] of boolean;
procedure bhf;
var
best,best_j:integer;
begin
fillchar(mark,sizeof(mark),false);
mark[1]:=true; b[1]:=0;{1为源点}
repeat
best:=0;
for i:=1 to n do
If mark then {对每一个已计算出最短路径的点}
for j:=1 to n do
if (not mark[j]) and (a[i,j]>0) then
if (best=0) or (b+a[i,j]<best) then begin
best:=b+a[i,j]; best_j:=j;
end;
if best>0 then begin
b[best_j]:=best;mark[best_j]:=true;
end;
until best=0;
end;{bhf}
B.Floyed算法求解所有顶点对之间的最短路径:
procedure floyed;
begin
for I:=1 to n do
for j:=1 to n do
if a[I,j]>0 then p[I,j]:=I else p[I,j]:=0; {p[I,j]表示I到j的最短路径上j的前驱结点}
for k:=1 to n do {枚举中间结点}
for i:=1 to n do
for j:=1 to n do
if a[i,k]+a[j,k]<a[i,j] then begin
a[i,j]:=a[i,k]+a[k,j];
p[I,j]:=p[k,j];
end;
end;
C. Dijkstra 算法:
var
a:array[1..maxn,1..maxn] of integer;
b,pre:array[1..maxn] of integer; {pre指最短路径上I的前驱结点}
mark:array[1..maxn] of boolean;
procedure dijkstra(v0:integer);
begin
fillchar(mark,sizeof(mark),false);
for i:=1 to n do begin
d:=a[v0,i];
if d<>0 then pre:=v0 else pre:=0;
end;
mark[v0]:=true;
repeat {每循环一次加入一个离1集合最近的结点并调整其他结点的参数}
min:=maxint; u:=0; {u记录离1集合最近的结点}
for i:=1 to n do
if (not mark) and (d<min) then begin
u:=i; min:=d;
end;
if u<>0 then begin
mark:=true;
for i:=1 to n do
if (not mark) and (a[u,i]+d<d) then begin
d:=a[u,i]+d;
pre:=u;
end;
end;
until u=0;
end;
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