並查集

做兩遍生成樹，第一遍優先選$1$邊求出必須選的$0$邊。第二遍先把必選的$0$邊選了，再選滿$k$條$0$邊就好了。

注意判$\text{no solution}$的三種情況：圖不連通或$0$邊個數$<k$或必選$0$邊個數$>k$。

程式碼：

#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 20005
#define M 100005
#define
 
 F inline
using namespace std;
struct edge{ int x,y,z,f; }ed[M];
int n,m,k,fa[N],ans[N];
F char readc(){
	static char buf[100000],*l=buf,*r=buf;
	if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
	return l==r?EOF:*l++;
}
F int _read(){
	int x=0; char ch=readc();
	while (!isdigit(ch)) ch=readc();
	while (isdigit(ch)
 
) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
	return x;
}
F bool cmp1(edge a,edge b){ return a.z>b.z; }
F bool cmp2(edge a,edge b){ return a.f==b.f?a.z<b.z:a.f>b.f; }
int findfa(int x){ return x==fa[x]?x:fa[x]=findfa(fa[x]); }
int main(){
	n=_read(),m=_read(),k=_read();
	for (int i=1,x,
 
y,f;i<=m;i++)
		x=_read(),y=_read(),f=_read(),ed[i]=(edge){x,y,f};
	for (int i=1;i<=n;i++) fa[i]=i;
	sort(ed+1,ed+m+1,cmp1); int s=0;
	for (int i=1,x,y;i<=m;i++)
		if ((x=findfa(ed[i].x))!=(y=findfa(ed[i].y)))
			fa[x]=y,ed[i].f=ed[i].z^1,s+=ed[i].f;
	if (s>k) return puts("no solution"),0;
	for (int i=2;i<=n;i++)
		if (findfa(fa[i])!=findfa(fa[1])) return puts("no solution"),0;
	for (int i=1;i<=n;i++) fa[i]=i;
	sort(ed+1,ed+m+1,cmp2),s=0;
	for (int i=1,x,y;i<=m;i++){
		if ((x=findfa(ed[i].x))==(y=findfa(ed[i].y))) continue;
		if (s==k&&ed[i].z==0) continue;
		fa[x]=y,ans[++ans[0]]=i,s+=(ed[i].z^1);
	}
	if (s<k) return puts("no solution"),0;
	for (int i=2;i<=n;i++)
		if (findfa(fa[i])!=findfa(fa[1])) return puts("no solution"),0;
	for (int i=1,j;i<n;i++)
		j=ans[i],printf("%d %d %d\n",ed[j].x,ed[j].y,ed[j].z);
	return 0;
}