【ZJOI2017 Round1练习&BZOJ4773】D3T1 cycle（最小负环，倍增）

题意：给定一个带权有向图，求点数最小的负环。

2 ⩽ n ⩽ 300
0 ⩽ m ⩽ n(n - 1)
1 ⩽ ui,vi ⩽ n
abs(w[j])<= 10^4

思路：倍增思想

设d[i,j,k]为走不多于2^i次步，从j走到k的最小权值和

显然d[i]可以由d[i-1]推出

f[i,j]表示当前走若干步后从i到j的最小权值和

从大到小枚举在原来的基础上再走不多于2^i步的结果

如果有负环就不用再走2^i步，将f复原

否则将j更新为走2^i步之后的数值，继续枚举

 const oo=;
 var d:array[..,..,..]of longint;
     g,f:array[..,..]of longint;
     n,m,i,j,k,t,ans,x,y,z:longint;
     flag:boolean;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 begin
  assign(input,'cycle.in'); reset(input);
  assign(output,'cycle.out'); rewrite(output);
  readln(n,m);
  for i:= to n do
   for j:= to n do
    if i<>j then d[,i,j]:=oo;
  for i:= to m do
  begin
   readln(x,y,z);
   d[,x,y]:=min(d[,x,y],z);
  end;
  for t:= to  do
  begin
   for i:= to n do
    for j:= to n do
     if i<>j then d[t,i,j]:=oo;
   for i:= to n do
    for j:= to n do
     for k:= to n do d[t,i,k]:=min(d[t,i,k],d[t-,i,j]+d[t-,j,k]);
  end;
  for i:= to n do
   for j:= to n do
    if i<>j then f[i,j]:=oo;
  for t:= downto  do
  begin
   for i:= to n do
    for j:= to n do g[i,j]:=f[i,j];
   for i:= to n do
    for j:= to n do
     for k:= to n do f[i,k]:=min(f[i,k],g[i,j]+d[t,j,k]);
   flag:=false;
   for i:= to n do
    if f[i,i]< then begin flag:=true; break; end;
   if flag then
   begin
    for i:= to n do
     for j:= to n do f[i,j]:=g[i,j];
   end
    else ans:=ans+(<<t);
  end;
  inc(ans);
  if ans>n then writeln()
   else writeln(ans);

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