UVA 10600 ACM Contest and Blackout

In order to prepare the �The First National ACM School Contest�(in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ). So, in order to do that, power station �Future� and one school (doesn�t matter which one) must be connected; in addition, some schools must be connected as well.

 

You may assume that a school has a reliable source of power if it�s connected directly to �Future�, or to any other school that has a reliable source of power. You are given the cost of connection between some schools. The major has decided to pick out two the cheapest connection plans � the cost of the connection is equal to the sum of the connections between the schools. Your task is to help the major � find the cost of the two cheapest connection plans.

 

Input

The Input starts with the number of test cases, T (1�T�15) on a line. Then T test cases follow. The first line of every test case contains two numbers, which are separated by a space, N (3�N�100) the number of schools in the city, and M the number of possible connections among them. Next M lines contain three numbers A_i, B_i, C_i , where C_i� is the cost of the connection (1�C_i�300) between schools A_i� and B_i. The schools are numbered with integers in the range 1 to N.

 

Output

For every test case print only one line of output. This line should contain two numbers separated by a single space - the cost of two the cheapest connection plans. Let S₁ be the cheapest cost and S₂ the next cheapest cost. It�s important, that S₁=S₂ if and only if there are two cheapest plans, otherwise S₁�S₂. You can assume that it is always possible to find the costs S₁ and S₂..

 

Sample Input

Sample Output

2

5 8

1 3 75

3 4 51

2 4 19

3 2 95

2 5 42

5 4 31

1 2 9

3 5 66

9 14

1 2 4

1 8 8

2 8 11

3 2 8

8 9 7

8 7 1

7 9 6

9 3 2

3 4 7

3 6 4

7 6 2

4 6 14

4 5 9

5 6 10

110 121

37 37

解析

次小生成树

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define INF 0x3f3f3f3fint Map[110][110],f[110][110],N,M,Dist[110],Pre[110];//int Fa[110],Deep[110];void readdata(){memset(Map,0x3f,sizeof(Map));scanf("%d%d",&N,&M);//for(int i=1;i<=N;i++) {Fa[i]=i;Deep[i]=1;}for(int i=1;i<=M;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);Map[a][b]=Map[b][a]=c;}}/*int find(int x){if(Fa[x]==x) return x;Fa[x]=find(Fa[x]);return Fa[x];}void merge(int a,int b);{int af=find(a),bf=find(b);if(deep[af]>deep[bf]) Fa[bf]=af;else if(deep[af]<deep[bf]) Fa[af]=bf;else {Fa[af]=bf;deep[bf]++;}}*/void prim(){Dist[1]=-1;for(int i=2;i<=N;i++) {Dist[i]=Map[i][1]; Pre[i]=1;}int sum=0;for(int i=2;i<=N;i++){int min=INF,k=0;for(int j=2;j<=N;j++)if(Dist[j]!=-1 && min>Dist[j]) {min=Dist[j];k=j;}if(k==0) break;sum+=min;Map[k][Pre[k]]=Map[Pre[k]][k]=INF;for(int j=1;j<=N;j++) if(Dist[j]==-1) f[k][j]=f[j][k]=max(min,f[Pre[k]][j]);Dist[k]=-1;for(int j=2;j<=N;j++)if(Dist[j]!=-1 && Dist[j]>Map[j][k]) {Dist[j]=Map[j][k];Pre[j]=k;}}printf("%d ",sum);int min=INF;for(int i=1;i<=N;i++)for(int j=1;j<=N;j++){if(i==j) continue;if(Map[i][j]!=INF && min>Map[i][j]-f[i][j])min=Map[i][j]-f[i][j];}printf("%d\n",sum+min);}int main(){freopen("ACMContestAndBlackout.in","r",stdin);freopen("ACMContestAndBlackout.out","w",stdout);int T;scanf("%d",&T);for(int i=1;i<=T;i++){readdata();prim();}return 0;}