COGS 497——奶牛派对

奶牛派对

我们发现每头牛需要走的路程即为它到x的最短路+x到它的最短路。

转化：

于是这道题变成了一道典型的单源最短路问题，只需求出每个点到x的最短路dl，以及从x到此点的最短路d2，然后去找max(dl+d2)即可。

效率分析：

使用dijsktra算法，时间复杂度为O(n^2)。

【我的程序】

 type aa=array[..,..] of longint;
 var
   n,m,x,a,b,t,i,j,k,min,max:longint;
   map1,map2:aa;
   ans:array[..] of longint;
   f:array[..] of boolean;
   d:array[..] of longint;
 procedure dijkstra(map:aa);
 begin
   for i:= to n do
    begin d[i]:=map[x,i]; f[i]:=false; end;
   f[x]:=true;
   for i:= to n do
    begin
      min:=maxlongint; k:=;
      for j:= to n do
       if (not f[j]) and (d[j]<min) then
        begin min:=d[j]; k:=j; end;
       if (k=) or (min=maxlongint) then exit;
      f[k]:=true;
      for j:= to n do
       if (not f[j]) and (d[k]+map[k,j]<d[j])
        then d[j]:=d[k]+map[k,j];
    end;
   for i:= to n do ans[i]:=ans[i]+d[i];
 end;
 begin
 assign(input,'party.in');
 reset(input);
 assign(output,'party.out');
 rewrite(output);
   readln(n,m,x);
   for i:= to n do
    for j:= to n do
     if i=j then begin map1[i,j]:=; map2[i,j]:=; end
      else begin map1[i,j]:=maxlongint div ; map2[i,j]:=maxlongint div ; end;
   for i:= to m do
    begin
      readln(a,b,t);
      map1[a,b]:=t;
      map2[b,a]:=t;//if (map[a,b]>t) or (map[a,b]=) then map[a,b]:=t;
    end;
    dijkstra(map1);
     dijkstra(map2);
    for i:= to n do if ans[i]>max then max:=ans[i];
    writeln(max);
 end.

【老师给的标程】

 const
  maxn=;
  maxm=;
 var
  n,m,X,i,j,sum,ans:longint;
  path:array[..,..maxn,..maxn]of Longint;
  dis:array[..,..maxn]of Longint;
  v:array[..maxn]of boolean;
 procedure Init;
 var
  A,b,c:Longint;
 begin
  readln(n,m,X);
  for i := to n do
     for j:= to n do
    begin
      path[,i,j]:=maxm;
    path[,i,j]:=maxm;
    end;
  for i :=  to m do
  begin
     readln(A,b,c);
  If path[,A,b] > c then
    begin
     path[,A,b] := c;
     path[,b,A] := c;
    end;
  end;
 end;
 procedure dij(k:longint);
 var
  min,minj,i,j,t:Longint;
 begin
  fillchar(v,sizeof(v),false);
  for i :=  To n do
     If path[k,X,i] <> maxm then dis[k,i] := path[k,X,i]
                            else dis[k,i] := maxm;
  v[x] := true;
  dis[k,x] := ;
  for i := n- downto  do
  begin
   min:=maxm;
   for j :=  to n do
   if (dis[k,j] < min)And(not v[j]) then
     begin
       min := dis[k,j];
       minj := j;
     end;
   if min<>maxm then
   begin
    v[minj] := true;
    j:=minj;
    for t :=  to n do
      If (dis[k,j]+path[k,j,t] < dis[k,t])And(not v[t]) Then
     dis[k,t] := dis[k,j]+path[k,j,t];
   end;
  end;
 end;
 procedure main;
 begin
  dij();
  dij();
  Ans:=;
  for i :=  to n do
   begin
     sum := dis[,i]+dis[,i];
    If (sum>ans)and(i<>X) then ans:=sum;
   end;
 end;
 procedure Ouit;
 begin
  Writeln(ans);
 end;
 begin
  Init;
  main;
  Ouit;
 end.

【考试时错误的做法】还是没有真正理解dijkstra

 var
   i,j,n,m,x,k1,k2,w,min,k,max:longint;
   a:array[..,..] of longint;
   d,ans:array[..] of longint;
   f:array[..] of boolean;
 procedure dijkstra1;
 begin
  for i:= to n do
  begin  d[i]:=a[x,i]; f[i]:=false; end;
  f[x]:=true;
   for i:= to n do
    begin
      min:=maxlongint;  k:=;
      for j:= to n do
      if (not f[j]) and (d[j]<min)  then
      begin
       min:=d[j]; k:=j;
      end;
    if  (k=)or(min=maxint) then exit;
     f[k]:=true;
     for j:= to n do
     if (not f[j]) and (d[k]+a[k,j]<d[j]) then d[j]:=d[k]+a[k,j];
    end;
 end;
 procedure dijkstra2;
 begin
  for i:= to n do
  begin  d[i]:=a[i,x]; f[i]:=false; end;
  f[x]:=true;
   for i:= to n do
    begin
      min:=maxlongint;  k:=;
      for j:= to n do
      if (not f[j]) and (d[j]<min)  then
      begin
       min:=d[j]; k:=j;
      end;
    if  (k=)or(min=maxint) then exit;
     f[k]:=true;
     for j:= to n do
     if (not f[j]) and (d[k]+a[k,j]<d[j]) then d[j]:=d[k]+a[k,j];
    end;
 end;
 begin
   readln(n,m,x);
   for i:= to n do
    for j:= to n do
     if i=j then a[i,j]:= else a[i,j]:=maxlongint div ;
   for i:= to m do
   begin
    readln(k1,k2,w);
    a[k1,k2]:=w;
   end;
    dijkstra1;
      for i:= to n do if i<>x then begin ans[i]:=ans[i]+d[i];  end;
    dijkstra2;     max:=-maxint;
      for i:= to n do if i<>x then
      begin
        ans[i]:=ans[i]+d[i];
         if ans[i]>max then max:=ans[i];
      end;
     writeln(max);
 end.