Connect them

You have n computers numbered from 1 to n and you want to connect them to make a small local area network (LAN). All connections are two-way (that is connecting computers i and j is the same as connecting computers j and i). The cost of connecting computer i and computer j is cij. You cannot connect some pairs of computers due to some particular reasons. You want to connect them so that every computer connects to any other one directly or indirectly and you also want to pay as little as possible.

Given n and each cij , find the cheapest way to connect computers.

Input

There are multiple test cases. The first line of input contains an integer T (T <= 100), indicating the number of test cases. Then T test cases follow.

The first line of each test case contains an integer n (1 < n <= 100). Then n

lines follow, each of which contains n integers separated by a space. The j-th integer of the i-th line in these n lines is cij, indicating the cost of connecting computers i and j (cij = 0 means that you cannot connect them). 0 <= cij <= 60000, cij = cji, cii = 0, 1 <= i, j <= n.

Output

For each test case, if you can connect the computers together, output the method in in the following fomat:

i1 j1 i1 j1 ......

where ik ik (k >= 1) are the identification numbers of the two computers to be connected. All the integers must be separated by a space and there must be no extra space at the end of the line. If there are multiple solutions, output the lexicographically smallest one (see hints for the definition of "lexicography small") If you cannot connect them, just output "-1" in the line.

Sample Input

2
3
0 2 3
2 0 5
3 5 0
2
0 0
0 0

Sample Output

1 2 1 3
-1

Hint

s:
A solution A is a line of p integers: a1, a2, ...ap.
Another solution B different from A is a line of q integers: b1, b2, ...bq.
A is lexicographically smaller than B if and only if:
(1) there exists a positive integer r (r <= p, r <= q) such that ai = bi for all 0 < i < r and ar < br
OR
(2) p < q and ai = bi for all 0 < i <= p

題目連結：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3204

題意描述：

給你n個點然後n*n條邊，如果權值為0則說明這兩點不通，問你能不能把所有的點連起來，如果能按字典序輸出這些點，如果不能則輸出-1，注意，在建立邊的時候需要按字典序，還有在最後輸出邊的時候也要按字典序。

解題思路：

首先把這些邊存到結構體裡，然後排序用克魯斯卡爾演算法，把要建立的邊存到另一個結構體裡，然後再次排序，輸出即可。

程式程式碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int inf=99999999;
struct data{
	int u;
	int v;
	int w;
}e[10010];
struct F{
	int a;
	int b;
}s[110];
int f[110];

int cmp(data x,data y);
int cmp1(F x,F y);
int getf(int u);
int merge(int u,int v);

int main()
{
	int n,T,i,j,m,t,count;
	scanf("%d",&T);
	while(T--)
	{
		t=1;
		count=0;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
			{
				scanf("%d",&m);
				e[t].u=i;
				e[t].v=j;
				e[t].w=m;
				if(m==0)
					e[t].w=inf;
				t++;
			}
		sort(e+1,e+t,cmp);
		for(i=1;i<=n;i++)
			f[i]=i;	
		for(i=1;i<t;i++)
		{
			if(merge(e[i].u,e[i].v)==1&&e[i].w<inf)
			{
				s[count].a=e[i].u;
				s[count].b=e[i].v;
				count++;
			}
			if(count==n-1)
				break;
		}	
		if(count!=n-1)
			printf("-1\n");
		else
		{
			sort(s,s+count,cmp1);
			for(i=0;i<count-1;i++)
				printf("%d %d ",s[i].a,s[i].b);
			printf("%d %d\n",s[count-1].a,s[count-1].b);
		}
	}
	return 0;
}
int cmp(data x,data y)
{
	if(x.w!=y.w)
		return x.w<y.w;
	else if(x.u!=y.u)
		return x.u<y.u;
	else
		return x.v<y.v;
}

int cmp1(F x,F y)
{
	if(x.a!=y.a)
		return x.a<y.a;
	return x.b<y.b;
}
int getf(int u)
{
	if(u==f[u])
		return u;
	f[u]=getf(f[u]);
	return f[u];
}

int merge(int u,int v)
{
	u=getf(u);
	v=getf(v);
	if(u!=v)
	{
		f[v]=u;
		return 1;
	}	
	return 0;
}