ZOJ

Connect them ZOJ - 3204

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 c_ij. 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 c_ij , 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 c_ij, indicating the cost of connecting computers iand j (c_ij = 0 means that you cannot connect them). 0 <= c_ij <= 60000, c_ij = c_ji, c_ii= 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 i_k i_k (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.

<b< dd="">

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


题意：最小生成树，但要求最小字典序的解。
那么变化就是在cmp的函数书写的，不能仅仅的比较value，还要把from，to按字典序的顺序来加入，当然输出也是如此。
所以重写两个cmp函数就可以了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=105;
int T,x;
int n,p,q;
int cnt,sum;
struct Node
{
    int from,to;
    double value;
}node[maxn*maxn],ans[maxn*maxn];
int fa[maxn];
bool cmp(Node a,Node b)
{
    if(a.value!=b.value)
        return a.value<b.value;
    else if(a.from!=b.from)
        return a.from<b.from;
    else
        return a.to<b.to;
}
bool cmpp(Node a,Node b)
{
    if(a.from==b.from)
        return a.to<b.to;
    else
        return a.from<b.from;
}
void init()
{
    for(int i=0;i<maxn;i++)
        fa[i]=i;
}
int findd(int x)
{
    if(fa[x]==x)
        return x;
    else
        return fa[x]=findd(fa[x]);
}
void Kruskal()
{
    sum=0;
    for(int i=1;i<=cnt;i++)
    {
        int fx=findd(node[i].from);
        int fy=findd(node[i].to);
        if(fx!=fy)
        {
            ans[sum++]=node[i];
            fa[fx]=fy;
        }
    }
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        cnt=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&x);
                if(x==0||j<=i)
                    continue;
                cnt++;
                node[cnt].from=i;node[cnt].to=j;
                node[cnt].value=x;
            }
        }
        init();
        sort(node+1,node+cnt+1,cmp);
        sum=0;
        Kruskal();
        if(sum!=n-1)
        {
            printf("-1
");
            continue;
        }
        else
        {
            sort(ans,ans+sum,cmpp);
            for(int i=0;i<sum-1;i++)
                printf("%d %d ",ans[i].from,ans[i].to);
            printf("%d %d
",ans[sum-1].from,ans[sum-1].to);
        }
    }
    return 0;
}