uva 10600 ACM Contest and Blackout(次小生成树)

Problem A

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 Ai, Bi, Ci , where Ci is the cost of the connection (1£Ci£300) between schools Ai and Bi.
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

Problem source: Ukrainian National Olympiad in Informatics 2001

Problem author: Shamil Yagiyayev

Problem submitter: Dmytro Chernysh

Problem solution: Shamil Yagiyayev, Dmytro Chernysh, K M Hasan

/*
一次过，，幸福来的太强烈了，加油！！！
今年在学校的最后一题。。。。
Time:2015-1-25 16:55
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAX=105;
bool vis[MAX],used[MAX][MAX];
int pre[MAX],dis[MAX],g[MAX][MAX],path[MAX][MAX];
int n,m;
int Prim(){
    for(int i=0;i<=n;i++){
        dis[i]=g[1][i];
        vis[i]=false;
        pre[i]=1;
    }
    int Mst=0;
    int minV,u;vis[1]=true;
    for(int i=1;i<=n;i++){
        u=1,minV=INF;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&minV>dis[j]){
                minV=dis[j];
                u=j;
            }
        }
        if(minV==INF) return Mst;
        Mst+=dis[u];vis[u]=true;
        used[u][pre[u]]=used[pre[u]][u]=true;
        for(int j=1;j<=n;j++){
            if(vis[j]&&j!=u){
                path[u][j]=path[j][u]=max(path[pre[u]][j],dis[u]);//此处是pre[u]到j，犹豫了一下
            }
            if(!vis[j]&&dis[j]>g[u][j]){
                dis[j]=g[u][j];
                pre[j]=u;
            }
        }
    }
    return Mst;
}
void Init(){
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            if(i==j)g[i][j]=0;
            else g[i][j]=g[j][i]=INF;

            path[i][j]=path[j][i]=0;
            used[i][j]=used[j][i]=false;
        }
    }

}
int main(){
    int T;
    int u,v,w;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        Init();
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&u,&v,&w);
            if(g[u][v]>w)
            g[u][v]=g[v][u]=w;
        }
        int Mst=Prim();
        int ans1,ans2=INF;
        ans1=Mst;
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                if(!used[i][j]){
                    ans2=min(ans2,Mst-path[i][j]+g[i][j]);
                }
            }
        }
        printf("%d %d\n",ans1,ans2);
    }
return 0;
}