hugedata

这个算法很简单，但是超时：var  n,m,i,x,y:longint;  f:array[1..1000,1..1000] of boolean;procedure floyd;var  k,i,j:longint;begin  for k:=1 to n do    for i:=1 to n do      forj:=1 to n do       f[i,j]:

var  a,b,x,y,d,e:longint;function extendedeuclid(a,b:longint;varx,y:longint):longint;var  t:longint;begin  if b=0 then    begin     x:=1;     y:=0;     exit(a);    end  else    begin

const  fx:array[1..4]of longint=(1,0,-1,0);  fy:array[1..4]of longint=(0,1,0,-1);var  map:array[0..5,0..5]of char;  ans,a,b,c,d:longint;procedure init;var  i,j:longint;  sss:char;begin  for

var  n,i,j,ans,cost:longint;  w:array[1..100,1..100] of longint;  d:array[1..100] of longint;  pre:array[1..100] of longint;  a,b:array[1..100] of longint;  used:array[1..100] of boolean;proce

var  n,m,i,a,b,s,t:longint;  x,y:array[1..100] of longint;  w:array[1..100,1..100] of real;  d:array[1..100] of real;  used:array[1..100] of boolean;procedure dijkstra(s:longint);var  i,k,pos:

var  n,m,i,x,y,j,ans:longint;  f:array[1..1000] of longint;  e:array[1..1000,0..1000] of longint;  used:array[1..1000] of boolean;  s:char;function findset(x:longint):longint;begin  if x<>f[x]

type  state=array[1..3,1..3] of integer;  node=record    x,y:integer;    dep:integer;    s:state;  end;const  t:state=((1,2,3),(8,0,4),(7,6,5));  dx:array[1..4]of integer=(1,-1,0,0);  dy:arr

var  n,m,a,b,i,max:longint;  w:array[1..300,1..300] of longint;  d:array[1..300] of longint;  used:array[1..300] of boolean;procedure prim(x:longint);var  i,k,min,pos:longint;begin  fillchar(

总的来说，这是一道考图论知识的题，可以搜到很多关于树的直径的东西，以及一些证明，这是一道好题！programshigong;type arr=array[1..200000]oflongint;varn,m,tot,{边数}   one,{以节点1为树根时dfs得到的最远点}  two:longint;{以one为树根时dfs得到的最远点，d(one,two)既为树的直径}

还须加强递归的理解啊。。。const  maxn=30000;var  n,i,x,y:longint;  h:array[1..maxn] of longint;{队首}  d:array[1..maxn] of longint;{距离队首的深度}  t:array[1..maxn] of longint;{后面接着的个数}  s:char;function findset(x