【CF666B】World Tour（贪心，最短路）

题意：给你一张有向图，叫你给出四个点的序列a,b,c,d，使得这四个点依次间的最短路之和最大。(4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000)

思路：O(n4)可用来对拍

我们需要O（n2）级别的算法

若枚举c,d，预处理出x到b比较远的3个x，d到y比较远的3个y，时间复杂度O(9n2)

为什么是3个而不是2个？

abc已枚举，若d的备选是ab就要GG，所以应该多一个备用，也就是三个点

 var c,d:array[..,..,..]of longint;
     head,vet,next,b,flag,q,x,y:array[..]of longint;
     dis:array[..]of longint;
     save:array[..,..]of longint;
     n,m,tot,i,s1,s2,s3,s4,a1,b1,c1,d1,ans,j,k,s,p1,q1:longint;

 procedure add(a,b:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  head[a]:=tot;
 end;

 procedure bfs(x:longint);
 var t,w,e,u,v:longint;
 begin
  fillchar(b,sizeof(b),);
  fillchar(dis,sizeof(dis),);
  t:=; w:=; q[]:=x; b[x]:=; dis[x]:=;
  while t<=w do
  begin
   u:=q[t]; inc(t);
   e:=head[u];
   while e<> do
   begin
    v:=vet[e];
    if b[v]= then
    begin
     dis[v]:=dis[u]+;
     b[v]:=;
     inc(w); q[w]:=v;
    end;
    e:=next[e];
   end;
  end;
 end;

 begin
  //assign(input,'1.in'); reset(input);
  //assign(output,'1.out'); rewrite(output);
  readln(n,m);
  for i:= to m do
  begin
   readln(x[i],y[i]);
   add(x[i],y[i]);
  end;

  for i:= to n do
  begin
   bfs(i);
   k:=; s:=;
   fillchar(flag,sizeof(flag),);
   for j:= to n do
    if (flag[j]=)and(dis[j]>s) then
    begin
     k:=j; s:=dis[j];
    end;
   flag[k]:=; c[i,,]:=k; c[i,,]:=s;
   k:=; s:=;
   for j:= to n do
    if (flag[j]=)and(dis[j]>s) then
    begin
     k:=j; s:=dis[j];
    end;
   flag[k]:=; c[i,,]:=k; c[i,,]:=s;
   k:=; s:=;
   for j:= to n do
    if (flag[j]=)and(dis[j]>s) then
    begin
     k:=j; s:=dis[j];
    end;
   flag[k]:=; c[i,,]:=k; c[i,,]:=s; // x---->c[i] ,,
   for j:= to n do save[i,j]:=dis[j];
  end;

  fillchar(head,sizeof(head),);
  tot:=;
  for i:= to m do add(y[i],x[i]);

  for i:= to n do
  begin
   bfs(i);
   k:=; s:=;
   fillchar(flag,sizeof(flag),);
   for j:= to n do
    if (flag[j]=)and(dis[j]>s) then
    begin
     k:=j; s:=dis[j];
    end;
   flag[k]:=; d[i,,]:=k; d[i,,]:=s;
   k:=; s:=;
   for j:= to n do
    if (flag[j]=)and(dis[j]>s) then
    begin
     k:=j; s:=dis[j];
    end;
   flag[k]:=; d[i,,]:=k; d[i,,]:=s;
   k:=; s:=;
   for j:= to n do
    if (flag[j]=)and(dis[j]>s) then
    begin
     k:=j; s:=dis[j];
    end;
   flag[k]:=; d[i,,]:=k; d[i,,]:=s; // d[i]    ---->i
  end;

  s1:=; s2:=; s3:=; s4:=; ans:=-maxlongint;
  for b1:= to n do
   for c1:= to n do
    if save[b1,c1]> then
    for p1:= to  do
     for q1:= to  do
     begin
      d1:=c[c1,p1,];
      a1:=d[b1,q1,];     //a1-->b1-->c1-->d1
      if (a1<>b1)and(a1<>c1)and(a1<>d1)and(b1<>c1)and(b1<>d1)and(c1<>d1) then
       if d[b1,q1,]+c[c1,p1,]+save[b1,c1]>ans then
       begin
        s1:=a1; s2:=b1; s3:=c1; s4:=d1;
        ans:=d[b1,q1,]+c[c1,p1,]+save[b1,c1];
       end;
     end;
  writeln(s1,' ',s2,' ',s3,' ',s4);

  //close(input);
  //close(output);
 end.