【luogu P3623 [APIO2008]免费道路】 题解

题目链接：https://www.luogu.org/problemnew/show/P3623
说是对克鲁斯卡尔的透彻性理解
正解：
先考虑加入水泥路，然后再考虑加入剩下必须要加入的最少鹅卵石路。
之后对原图再跑最小生成树
先跑鹅卵石路到k条。
再从所有水泥路中直到成为最小生成树。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
int n, m, k, fa[maxn], cnt, shuini, eluan, tot;
bool f[maxn];
struct edge{
	int u, v, w;
}e[maxn<<2], ans[maxn<<2];
bool cmp(edge a, edge b)
{
	return a.w < b.w;
}
int find(int x)
{
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i = 1; i <= n; i++) fa[i] = i;
	for(int i = 1; i <= m; i++)
	{
		int opt;
		scanf("%d%d%d",&e[i].u, &e[i].v, &opt);
		if(opt == 0) e[i].w = 1, eluan++;
		else e[i].w = 0, shuini++;
	}
	if(eluan < k)
	{
		cout<<"no solution
";
		return 0;
	}
	sort(e+1, e+1+m, cmp);
	for(int i = 1; i <= shuini; i++)
	{
		if(cnt == n-1) break;
		int rx, ry;
		rx = find(e[i].u), ry = find(e[i].v);
		if(rx != ry)
		{
			fa[ry] = rx;
			cnt++;
		}
	}
	int eluanmust = 0;
	for(int i = shuini+1; i <= m; i++)
	{
		if(cnt == n-1) break;
		int rx, ry;
		rx = find(e[i].u), ry = find(e[i].v);
		if(rx != ry)
		{
			fa[ry] = rx;
			f[i] = 1, eluanmust++;
			cnt++;
		}
	}
	if(cnt != n-1 || eluanmust > k) 
	{
		cout<<"no solution
";
		return 0;
	}
	for(int i = 1; i <= n; i++) fa[i] = i; 
	cnt = 0;
	for(int i = shuini+1; i <= m; i++)
	{
		if(cnt == n-1) break;
		if(f[i] == 1) 
		{
			int rx, ry;
			rx = find(e[i].u), ry = find(e[i].v);		
			fa[ry] = rx;
			cnt++;
			ans[++tot].u = e[i].u;
			ans[tot].v = e[i].v;
			ans[tot].w = e[i].w^1;
			//printf("%d %d %d
",e[i].u, e[i].v, e[i].w^1);
		}
	}
	for(int i = shuini+1; i <= m; i++)
	{
		if(cnt == k) break;
		int rx, ry;
		rx = find(e[i].u), ry = find(e[i].v);
		if(rx != ry)
		{
			fa[ry] = rx;
			cnt++;
			ans[++tot].u = e[i].u;
			ans[tot].v = e[i].v;
			ans[tot].w = e[i].w^1;
			//printf("%d %d %d
",e[i].u, e[i].v, e[i].w^1);
		}
	}
	if(cnt < k) 
	{
		cout<<"no solution
";
		return 0;
	}
	for(int i = 1; i <= shuini; i++)
	{
		if(cnt == n-1) break;
		int rx, ry;
		rx = find(e[i].u), ry = find(e[i].v);
		if(rx != ry)
		{
			fa[ry] = rx;
			cnt++;
			ans[++tot].u = e[i].u;
			ans[tot].v = e[i].v;
			ans[tot].w = e[i].w^1;
			//printf("%d %d %d
",e[i].u, e[i].v, e[i].w^1);
		}
	}
	if(cnt != n-1)
	{
		cout<<"no solution
";
		return 0;
	}
	for(int i = 1; i <= tot; i++)
	printf("%d %d %d
",ans[i].u, ans[i].v, ans[i].w);
	return 0;
}