ZOJ 3204 Connect them（最小生成树：kruscal算法）

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

Connect them

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Time Limit: 1 Second     Memory Limit:32768 KB

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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 computersi andj is the same as connecting computersj and i). The cost of connecting computeri and computerj 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 integerT (T <= 100), indicating the number of test cases. ThenT test cases follow.

The first line of each test case contains an integern (1 <n <= 100). Thenn lines follow, each of which containsn integers separated by a space. Thej-th integer of thei-th line in thesen lines is c_ij, indicating the cost of connecting computersi andj (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.

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, ...a_p.
Another solution B different from A is a line of q integers:b₁,b₂, ...b_q.
A is lexicographically smaller than B if and only if:
(1) there exists a positive integer r (r <= p, r <=q) such thata_i =b_i for all 0 <i < r anda_r <b_r
OR
(2) p < q and a_i = b_i for all 0 <i <=p

题目大意：题目大意：有几台电脑，怎么用最少的费用把他

           们连接起来

　　 

解题思路：最小生成树问题。先将边从小到大排序，依次把

           边两端的点用并查集合并。需要注意的是，题目中

           给出的矩阵上三角和下三角是相同的，即只需处理

           一半就可以啦。

<span style="font-size:24px;">///知识点：最小生成树
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 using namespace std;

 const int maxn=110;
 int f[maxn];
 struct Edge
 {
     int from,to;
     int w;
 }edge[maxn*maxn];
 int tol;
 Edge ans[maxn*maxn];
 int cnt;
 void addedge(int u,int v,int w)
 {
     edge[tol].from=u;
     edge[tol].to=v;
     edge[tol].w=w;
     tol++;
 }
 bool cmp1(Edge a,Edge b)
 {
     if(a.w!=b.w) return a.w<b.w;
     else if(a.from!=b.from) return a.from<b.from;
     else return a.to<b.to;
 }
 bool cmp2(Edge a,Edge b)
 {
     if(a.from!=b.from) return a.from<b.from;
     else return a.to<b.to;
 }
 int find(int x)
 {
     if(f[x]==-1) return x;
     return f[x]=find(f[x]);
 }
 void kruscal()
 {
     memset(f,-1,sizeof(f));
     cnt=0;
     for(int k=0;k<tol;k++)
     {
         int u=edge[k].from;
         int v=edge[k].to;
         int t1=find(u);
         int t2=find(v);
         if(t1!=t2)
         {
             ans[cnt++]=edge[k];
             f[t1]=t2;
         }
     }
 }
 int main()
 {
     int T;
     scanf("%d",&T);
     int n;
     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);
         }
         sort(edge,edge+tol,cmp1);
         kruscal();
         if(cnt!=n-1)
         {
             printf("-1\n");
             continue;
         }
         else
         {
             sort(ans,ans+cnt,cmp2);
             for(int i=0;i<cnt-1;i++)
                printf("%d %d ",ans[i].from,ans[i].to);
             printf("%d %d\n",ans[cnt-1].from,ans[cnt-1].to);
         }
     }
     return 0;
 }</span>