【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[1..3000,1..3,1..2]of longint;
    head,vet,next,b,flag,q,x,y:array[1..10000]of longint;
    dis:array[1..3000]of longint;
    save:array[1..3000,1..3000]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),0);
 fillchar(dis,sizeof(dis),0);
 t:=1; w:=1; q[1]:=x; b[x]:=1; dis[x]:=0;
 while t<=w do
 begin
  u:=q[t]; inc(t);
  e:=head[u];
  while e<>0 do
  begin
   v:=vet[e];
   if b[v]=0 then
   begin
    dis[v]:=dis[u]+1;
    b[v]:=1;
    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:=1 to m do
 begin
  readln(x[i],y[i]);
  add(x[i],y[i]);
 end;

 for i:=1 to n do
 begin
  bfs(i);
  k:=1; s:=0;
  fillchar(flag,sizeof(flag),0);
  for j:=1 to n do
   if (flag[j]=0)and(dis[j]>s) then
   begin
    k:=j; s:=dis[j];
   end;
  flag[k]:=1; c[i,1,1]:=k; c[i,1,2]:=s;
  k:=1; s:=0;
  for j:=1 to n do
   if (flag[j]=0)and(dis[j]>s) then
   begin
    k:=j; s:=dis[j];
   end;
  flag[k]:=1; c[i,2,1]:=k; c[i,2,2]:=s;
  k:=1; s:=0;
  for j:=1 to n do
   if (flag[j]=0)and(dis[j]>s) then
   begin
    k:=j; s:=dis[j];
   end;
  flag[k]:=1; c[i,3,1]:=k; c[i,3,2]:=s; // x---->c[i] 1,2,3
  for j:=1 to n do save[i,j]:=dis[j];
 end;

 fillchar(head,sizeof(head),0);
 tot:=0;
 for i:=1 to m do add(y[i],x[i]);

 for i:=1 to n do
 begin
  bfs(i);
  k:=1; s:=0;
  fillchar(flag,sizeof(flag),0);
  for j:=1 to n do
   if (flag[j]=0)and(dis[j]>s) then
   begin
    k:=j; s:=dis[j];
   end;
  flag[k]:=1; d[i,1,1]:=k; d[i,1,2]:=s;
  k:=1; s:=0;
  for j:=1 to n do
   if (flag[j]=0)and(dis[j]>s) then
   begin
    k:=j; s:=dis[j];
   end;
  flag[k]:=1; d[i,2,1]:=k; d[i,2,2]:=s;
  k:=1; s:=0;
  for j:=1 to n do
   if (flag[j]=0)and(dis[j]>s) then
   begin
    k:=j; s:=dis[j];
   end;
  flag[k]:=1; d[i,3,1]:=k; d[i,3,2]:=s; // d[i] 1 2 3 ---->i
 end;

 s1:=1; s2:=2; s3:=3; s4:=4; ans:=-maxlongint;
 for b1:=1 to n do
  for c1:=1 to n do
   if save[b1,c1]>0 then
   for p1:=1 to 3 do
    for q1:=1 to 3 do
    begin
     d1:=c[c1,p1,1];
     a1:=d[b1,q1,1];     //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,q1,2]+c[c1,p1,2]+save[b1,c1]>ans then
      begin
       s1:=a1; s2:=b1; s3:=c1; s4:=d1;
       ans:=d[b1,q1,2]+c[c1,p1,2]+save[b1,c1];
      end;
    end;
 writeln(s1,' ',s2,' ',s3,' ',s4);

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