bzoj1202

很久以前写的，忘补解题报告了
首先似乎dfs就可以了吧？
但还有更高大上的做法
其实这东西就是告诉sum[y]-sum[x-1]=z
然后给出一堆看成不成立
可以用并查集，维护每个点到father点的差即可

var sum,fa:array[0..1010] of longint;
    i,t,n,m,x,y,z,k1,k2:longint;
    ch:boolean;
function getf(x:longint):longint;
  var k:longint;
  begin
    if fa[x]=x then exit(x)
    else begin
      k:=fa[x];
      fa[x]:=getf(fa[x]);
      sum[x]:=sum[x]+sum[k];
      exit(fa[x]);
    end;
  end;

begin
  readln(t);
  while t>0 do
  begin
    readln(n,m);
    for i:=0 to n do
      fa[i]:=i;
    ch:=true;
    fillchar(sum,sizeof(sum),0);
    for i:=1 to m do
    begin
      readln(x,y,z);
      if not ch then continue;
      dec(x);
      k1:=getf(x);
      k2:=getf(y);
      if (k1<>k2) then
      begin
        fa[k1]:=k2;
        sum[k1]:=sum[y]-sum[x]+z;
      end
      else if sum[x]-sum[y]<>z then
      begin
        ch:=false;
        break;
      end;
    end;
    if ch then writeln('true') else writeln('false');
    dec(t);
  end;
end.