简单最小生成树，畅通project。这三道题目都是练习最小生成树的。

注意一下推断是否有通路时，kruskal能够推断每一个点的祖先是否同样。prim能够推断每一个点是否都加进集合里面了，也就是说是否都訪问过。

prim算法要把没有给的边初始化为MAX无穷大。

。。

代码：（kruskal）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<algorithm>
#include<utility>
#include<functional>
#define MAX 0x7fffffff

using namespace std;
int p[105];
struct node
{
	int i,j,len;
}gra[10005];
int find(int x)
{
	return x == p[x]? x: p[x] = find(p[x]);
}
int cmp(const void *a,const void *b)
{
	return ((node *)a)->len - ((node *)b)->len;
}
int m,n;
void kruskal()
{
	int i,sum = 0;
	for(i=1; i<=m; i++)
	{
		int x = find(gra[i].i);
		int y = find(gra[i].j);
		if(x != y)
		{
			sum += gra[i].len;
			p[x] = y;
		}
	}
	int flag = 0;
	for(i=1; i<=n; i++)
	{
		if(find(1) != find(i))
		{
			flag = 1;
			break;
		}
	}
	if(flag)
		cout << "?
" << endl;
	else
		cout << sum << endl;
	return ;
}
int main()
{
	int i,j,a,b,c;
	while(cin >> m >> n,m)
	{
		for(i=1; i<=m; i++)
		{
			cin >> a >> b >> c;
			gra[i].i = a;
			gra[i].j = b;
			gra[i].len = c;
		}
		for(i=1; i<=n; i++)
			p[i] = i;
		qsort(gra+1,m,sizeof(gra[0]),cmp);
		kruskal();
	} 
	return 0;
}

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<algorithm>
#include<utility>
#include<functional>
#define MAX 0x7fffffff

using namespace std;
int gra[105][105];
int m,n;
void prim()
{
	int visit[105],i,j,now,MIN,sum = 0;
	int d[105];
	memset(visit,0,sizeof(visit));
	for(i=1; i<=n; i++)
		d[i] = MAX;
	d[1] = 0;
	visit[1] = 1,now = 1;
	for(i=2; i<=n; i++)
	{
		for(j=1; j<=n; j++)
			if(!visit[j] && d[j]>gra[now][j])
				d[j] = gra[now][j];
		MIN = MAX;
		for(j=1; j<=n; j++)
			if(!visit[j] && MIN > d[j])
					MIN = d[now = j];
		visit[now] = 1;
		sum += d[now]; 
	}
	int flag = 0;
	for(i=1; i<=n; i++)
	{
		if(!visit[i])
		{
			flag = 1;
			break;
		}
	}
	if(flag)
		cout << "?" << endl;
	else
		cout << sum << endl;
}
int main()
{
	int i,j,a,b,c;
	while(cin >> m >> n,m)
	{
		for(i=1; i<=n; i++)
			for(j=1; j<=n; j++)
				gra[i][j] = MAX;
		for(i=1; i<=m; i++)
		{
			cin >> a >> b >> c;
			gra[a][b] = gra[b][a] = c;
		}
		prim();
	}
	return 0;
}