【Vijos1250】最勇敢的机器人（并查集，分组背包DP）

题意：有N个物品，承重上限为M，有K组物品互斥关系，互斥关系有传递性，即1与2互斥，2与3互斥，1与3也互斥

给出每个物品的花费和价值，求承重上限内的最大价值总和

n<=1000,m<=1000,k<=1000

c[i]<=1000 w[i]<=10

思路：日常刷水

因为互斥关系有传递性，并查集后就是分组背包DP

 var dp:array[..]of longint;
     a,c,w,f:array[..]of longint;
     n,m,i,j,k,ans,x,y,p,q:longint;

 function max(x,y:longint):longint;
 begin
  if x>y then exit(x);
  exit(y);
 end;

 function find(k:longint):longint;
 begin
  if f[k]<>k then f[k]:=find(f[k]);
  find:=f[k];
 end;

 begin
  assign(input,'Vijos1250.in'); reset(input);
  assign(output,'Vijos1250.out'); rewrite(output);

  readln(n,m,k);

  for i:= to n do read(c[i],w[i]);
  for i:= to n do f[i]:=i;
  for i:= to k do
  begin
   readln(x,y);
   p:=find(x); q:=find(y);
   if p<>y then f[q]:=p;
  end;
  for i:= to n do
  begin
   q:=;
   for j:= to n do
    if find(j)=i then begin inc(q); a[q]:=j; end;
   for j:=m downto  do
    for k:= to q do
     if j-w[a[k]]>= then dp[j]:=max(dp[j],dp[j-w[a[k]]]+c[a[k]]);
  end;
  ans:=;
  for i:= to m do ans:=max(ans,dp[i]);
  writeln(ans);

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