单对顶点间的最短路径 例1:输入起点,终点,求其最短路径?
1.下面是Dijkstra 算法: 基本思想是:设置一个顶点的集合s,并不断地扩充这个集合,一个顶点属于集合s当且仅当从源点到该点的路径已求出。开始时s中仅有源点,并且调整非s中点的最短路径长度,找当前最短路径点,将其加入到集合s,直到终点在s中。 programzudlouj;
const n=6;max=10000 ;
cost:array[1..6,1..6] ofreal=((0,50,10,max,45,max),(max,0
,15,max,10,max),(20,max,0,15,max,max),(max,20,max,0,35,max),(
max,max,max,30,0,max),(max,max,max,3,max,0));
var dist:array[1..n] of real;
path,p:array[1..n] of 1..n;
first,tail,u:1..n;
s:set of 1..n;
i,j,y,m:integer;
min:real;
begin
read(first,tail);
for i:=1 to n do dist:=max;
dist[first]:=0;
s:=[first]; u:=first;
while u<>tail do
begin
for j:= 1 to n do
if not(j in s) and (dist+cost[u,j]<dist[j]) then
begin dist[j]:=dist+cost[u,j];path[j]:=u end;
min:=max;
for j:=1 to n do
if not(j in s) and (dist[j]<min) then beginu:=j;min:=dist[j];end;
if min=max then begin writeln('No answer');halt end;
s:=s+;
end;
writeln('mindist(',first,',',tail,')=',dist[tail]:8:2);
y:=tail;m:=0;
while (y<>first) do
begin inc(m);p[m]:=y;y:=path[y]; end;
write('path:',first);
for j:=m downto 1 do
write('->',p[j]);
writeln;
end.
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