Zoj 3204 connect them (c) Minimum Spanning Tree Kruskal

Connect them Time Limit: 1 second memory limit: 32768 KB

You haveNComputers numbered from 1NAnd you want to connect them to make a small local area network (LAN). All connections are two-way (that is connecting computersIAndJIs the same as connecting computersJAndI). The cost of connecting computerIAnd computerJIsCIJ. You cannot connect some pairs of computers due to some special 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.

GivenNAnd eachCIJ, 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. ThenTTest Cases follow.

The first line of each test case contains an integerN(1 <N<= 100). ThenNLines follow, each of which containsNIntegers separated by a space.J-Th INTEGER OFI-Th line in theseNLines isCIJ, Indicating the cost of connecting computersIAndJ(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 the following fomat:

I1J1I1J1 ......

WhereIk 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, outputLexicographically smallestOne (see hints for the definition"Lexicography small") If you cannot connect them, just output"-1 "in the line.

Sample Input

 

230 2 32 0 53 5 020 00 0

 

Sample output

 

1 2 1 3-1

 

Hints:
A solutionAIs a linePIntegers:A1,A2,...AP.
Another solutionBDifferent fromAIs a lineQIntegers:B1,B2,...BQ.
AIsLexicographically smallerThanBIf and only if:
(1) there exists a positive integerR(R<=P,R<=Q) Such thatAI=BiFor all 0 <I<RAndAr<BR 
Or
(2)P<QAndAI=BiFor all 0 <I<=P

 

Minimum Spanning Tree. If not, output-1. If yes

The lexicographic arrangement of the output edge.
After wa several times, you must first sort the smallest Spanning Tree by lexicographically. Otherwise, the weights may be the same, but if you select a vertex with a large number, the second sorting will be useless.

 1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <string> 5 #include <vector> 6 #include <algorithm> 7 using namespace std; 8 int T, n, cnt, Count; 9 #define maxn 110*11010 struct Node{11     int u, v, w;12 }node[maxn];13 struct Point{14     int x, y;15 }point[110];16 int mp[110][110], pre[110];17 int find(int x){18     if(x == pre[x]) return x;19     else return find(pre[x]);20 }21 /*22 bool cmp(Node x, Node y){23     return x.w < y.w;24 }*/25 bool cmp(Node x, Node y){26     if(x.w == y.w){27         if(x.u == y.u) return x.v < y.v;28         else return x.u < y.u;29     }30     return x.w < y.w;31 }32 bool cmp2(Point a, Point b){33     if(a.x == b.x) return a.y < b.y;34     else return a.x < b.x;35 }36 int main(){37     scanf("%d", &T);38     while(T--){39         scanf("%d", &n);40         for(int i = 1; i <= n; i++){41             for(int j = 1; j <= n; j++) scanf("%d", &mp[i][j]);42         }43         for(int i = 1; i <= n; i++) pre[i] = i;44         cnt = 0;45         for(int i = 1; i <= n; i++){46             for(int j = 1; j <= n; j++){47                 if(mp[i][j] != 0 && (i<j)){48                     cnt++;49                     node[cnt].u = i; node[cnt].v = j; node[cnt].w = mp[i][j];50                 }51             }52         }53         Count = 0;54         sort(node+1, node+1+cnt, cmp);55         vector < Point > v;56         for(int i = 1; i <= cnt; i++){57             int aa = find(node[i].u);58             int bb = find(node[i].v);59             if(aa != bb){60                 pre[aa] = bb;61                 Point temp; temp.x = node[i].u; temp.y = node[i].v;62                 v.push_back(temp);63                 Count++;64             }65             if(Count == n-1) break;66         }67         if(Count != n-1) printf("-1\n");68         else{69             sort(v.begin(), v.end(), cmp2);70             for(int i = 0; i < v.size(); i++){71                 if(i == 0) printf("%d %d", v[i].x, v[i].y);72                 else printf(" %d %d", v[i].x, v[i].y);73             }74             printf("\n");75         }76         77         78     }79     80     return 0;81 }