【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[0..1000]of longint;
    a,c,w,f:array[1..1000]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:=1 to n do read(c[i],w[i]);
 for i:=1 to n do f[i]:=i;
 for i:=1 to k do
 begin
  readln(x,y);
  p:=find(x); q:=find(y);
  if p<>y then f[q]:=p;
 end;
 for i:=1 to n do
 begin
  q:=0;
  for j:=1 to n do
   if find(j)=i then begin inc(q); a[q]:=j; end;
  for j:=m downto 0 do
   for k:=1 to q do
    if j-w[a[k]]>=0 then dp[j]:=max(dp[j],dp[j-w[a[k]]]+c[a[k]]);
 end;
 ans:=0;
 for i:=0 to m do ans:=max(ans,dp[i]);
 writeln(ans);

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