ZOJ 3204 Connect them（最小生成树之Krusal 输出字典序最小的）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3367

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 thei-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

Hints:

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

Author: CAO, Peng

Source: The 6th Zhejiang Provincial Collegiate Programming Contest

题意：

输出最小生成树中字典序最小的！

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 147
int F[maxn];
struct Edge
{
    int u, v;
    int w;
} edge[maxn*maxn], ans[maxn*maxn];
int n;
int tol;
int cnt;
void addedge(int u,int v,int w)
{
    edge[tol].u = u;
    edge[tol].v = v;
    edge[tol].w = w;
    tol++;
}
bool cmp1(Edge a,Edge b)//保证按字典序从小输出
{
    if(a.w != b.w)
        return a.w < b.w;
    else if(a.u != b.u)
        return a.u < b.u;
    else
        return a.v < b.v;
}
bool cmp2(Edge a,Edge b)
{
    if(a.u != b.u)
        return a.u < b.u;
    else
        return a.v < b.v;
}
int find(int x)
{
    return x == F[x] ? x: F[x]=find(F[x]);
}
void kruscal()
{
    for(int i = 0; i <= n; i++)
    {
        F[i] = i;
    }
    sort(edge,edge+tol,cmp1);//对边的排序
    cnt = 0;
    for(int i = 0; i < tol; i++)
    {
        int u = edge[i].u;
        int v = edge[i].v;
        int t1 = find(u);
        int t2 = find(v);
        if(t1 != t2)
        {
            ans[cnt++] = edge[i];
            F[t1] = t2;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        tol = 0;
        int w;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
            {
                scanf("%d",&w);
                if(j <= i)continue;
                if(w == 0)continue;
                addedge(i,j,w);
            }

        kruscal();
        if(cnt!=n-1)
        {
            printf("-1\n");
        }
        else
        {
            sort(ans,ans+cnt,cmp2);
            for(int i = 0; i < cnt-1; i++)
                printf("%d %d ",ans[i].u,ans[i].v);
            printf("%d %d\n",ans[cnt-1].u,ans[cnt-1].v);
        }
    }
    return 0;
}