bzoj 3624 贪心+Kruscal（并查集）

题意：n个点，m条无向边。共有两类边，分别是0类边和1类边，要求找出一棵最小生成树满足其中0类边恰好有K条，如果无解输出“no solution”

实际上就是一道经典的特殊的MST

跑两边Kruscal

第一遍：优先选择1类边，即在保证是树的前提下，能用1类边就用1类边，然后剩下的用0类边补全，则这些0类边是必须要选的，如果必选的0类边的数量大于K，则无解

第二遍：先加入上一次必须要选的0类边，然后在保证是树的前提下，在剩下的0类边里选取来补足K条，如果最后选择的0类边总数依旧小于K条，则无解，否则用1类边来补全最小生成树

type
rec=record
x,y,c:longint;
end;

var
n,m,k,tt,tx,ty,t1:longint;
i :longint;
father :array[0..20010] of longint;
l :array[0..100010] of rec;
flag :array[0..100010] of boolean;

function get_father(x:longint):longint;
begin
if x=father[x] then exit(x);
father[x]:=get_father(father[x]);
exit(father[x]);
end;

begin
read(n,m,k);
for i:=1 to n do father[i]:=i;
for i:=1 to m do read(l[i].x,l[i].y,l[i].c);
tt:=0;
// use 1 as possible as you can
for i:=1 to m do
if l[i].c=1 then
begin
tx:=get_father(l[i].x);
ty:=get_Father(l[i].y);
if tx<>ty then
begin
father[tx]:=ty; inc(tt);
if tt=n-1 then break;
end;
end;
//find then necessary 0
for i:=1 to m do
if (l[i].c=0) and (tt<n-1) then
begin
tx:=get_father(l[i].x);
ty:=get_father(l[i].y);
if tx<>ty then
begin
inc(t1); inc(tt); father[tx]:=ty; flag[i]:=true;
if tt=n-1 then break;
end;
end;
//you have to use t1 0
if t1>k then
begin
writeln('no solution');exit;
end;
for i:=1 to n do father[i]:=i; tt:=0;
//use the necessary 0
for i:=1 to m do
if flag[i] then
begin
tx:=get_father(l[i].x);
ty:=get_father(l[i].y);
if tx<>ty then
begin
dec(k); inc(tt);
father[tx]:=ty;
if tt=n-1 then break;
end;
end;
//ensure use k 0
for i:=1 to m do
if (l[i].c=0) and not flag[i] and (tt<n-1) then
begin
tx:=get_father(l[i].x);
ty:=get_father(l[i].y);
if tx<>ty then
begin
dec(k); inc(tt);
father[tx]:=ty;
flag[i]:=true;
if (k=0) or (tt=n-1) then break;
end;
end;
//the number of 0 is less than k
if k>0 then
begin
writeln('no solution');exit;
end;
//find other 1 to make it complete
for i:=1 to m do
if (l[i].c=1) and (tt<n-1) then
begin
tx:=get_father(l[i].x);
ty:=get_father(l[i].y);
if tx<>ty then
begin
inc(tt);
father[tx]:=ty;
flag[i]:=true;
if tt=n-1 then break;
end;
end;
//
for i:=1 to m do if flag[i] then writeln(l[i].x,' ',l[i].y,' ',l[i].c);
end.——by Eirlys