Air Ports&&http://www.lightoj.com/volume_showproblem.php?problem=1059

最小生成树变形，这一题，真他妈的恶心，由于没看清最后一句话，导致一直wa，，在修建机场和修路方面，如果修路的费用和修飞机场的相同，则优先考虑修飞机场，，

法一：

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#define pf printf
#include<cstdio>
#define N 100005
#define M 10005
using namespace std;
typedef struct node
{
	int x;
	int y;
	int len;
	bool operator<(node a)const
	{return len<a.len;}
}Node;
Node s[N];
int Father[M];
int n,m,cost;
int Find(int x)
{
	if(x==Father[x]) return x;
	return Father[x]=Find(Father[x]);
}
void in(int &a)
{
	char ch;
	while((ch=getchar())<'0'||ch>'9');
	for(a=0;ch>='0'&&ch<='9';ch=getchar()) a=a*10+ch-'0';
}
int main()
{
	int T;
	in(T);
	for(int k=1;k<=T;++k)
	{
		in(n),in(m),in(cost);
		for(int i=0;i<=n;++i) Father[i]=i;
		for(int i=0;i!=m;++i) in(s[i].x),in(s[i].y),in(s[i].len);
		sort(s,s+m);
		int num=0;
		bool flag=false;
		long long ans=0;
		for(int i=0;i<m;++i)
		{
			if(s[i].len>=cost) continue;
			if(num==n-1) {flag=1;break;}
			int x=Find(s[i].x);
			int y=Find(s[i].y);
				if(x!=y)
				{
					num++;
					Father[x]=y;
					ans+=s[i].len;
				}
	     }
		if(flag) pf("Case %d: %lld 1\n",k,ans+cost);
		else
		{
			int res=0;
			for(int i=1;i<=n;++i)
				if(i==Find(i)) res++;
			pf("Case %d: %lld %d\n",k,ans+res*cost,res);

		}
	}return 0;
}

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#define pf printf
#include<cstdio>
#define N 100005
#define M 10005
using namespace std;
typedef struct node
{
	int x;
	int y;
	int len;
	bool operator<(node a)const
	{return len<a.len;}
}Node;
Node s[N];
int Father[M];
int n,m,cost;
int Find(int x)
{
	if(x==Father[x]) return x;
	return Father[x]=Find(Father[x]);
}
void in(int &a)
{
	char ch;
	while((ch=getchar())<'0'||ch>'9');
	for(a=0;ch>='0'&&ch<='9';ch=getchar()) a=a*10+ch-'0';
}
int main()
{
	int T;
	in(T);
	for(int k=1;k<=T;++k)
	{
		in(n),in(m),in(cost);
		for(int i=0;i<=n;++i) Father[i]=i;
		for(int i=0;i!=m;++i) in(s[i].x),in(s[i].y),in(s[i].len);
		sort(s,s+m);
		int num=0;
		long long ans=0;
		 for(int j=0;j<m;j++)
	    {
	     if(num==n-1) break;
		 if(s[j].len>=cost) continue;
	     int xx=Find(s[j].x),yy=Find(s[j].y);
	     if(xx!=yy)
	     {
			 num++;
	         Father[xx]=yy;
		     ans+=s[j].len;
		 }
	 }
	 printf("Case %d: %lld %d\n",k,(ans+(n-num)*cost),n-num);
	}return 0;
}