UVA 10600 ACM Contest and Blackout 次小生成树/裸

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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 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

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

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一个裸的次小生成树，没什么好说的，我肯定做麻烦了

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=211;
const int maxm=21111;
const double eps=1e-12;
const double OO=1e100;

struct Road{
    int u,v;
    int w;
    Road(){}
    Road(int u,int v,int w){
        this->u=u;
        this->v=v;
        this->w=w;
    }
    bool operator<(const Road& rhs) const{
        return w<rhs.w;
    }
};

class DisjointSet{
private:
    int pa[maxn];
    int n;
public:
    void makeSet(int n){
        this->n=n;
        for (int i=0;i<=n;i++) pa[i]=i;
    }
    int findSet(int x){
        if (x!=pa[x]) pa[x]=findSet(pa[x]);
        return pa[x];
    }
    void unionSet(int x,int y){
        x=findSet(x);
        y=findSet(y);
        if (x!=y) pa[x]=y;
    }
}disjointSet;

class Graph{
private:
    struct EdgeNode{
        int to;
        int w;
        int next;
    };
    int head[maxn],edge,n;
    EdgeNode edges[maxm*2];
    int maxCost[maxn][maxn];
    void dfsCost(int s,int u,int pa){
        for (int i=head[u];i!=-1;i=edges[i].next){
            int v=edges[i].to;
            int w=edges[i].w;
            if (v==pa) continue;
            maxCost[s][v]=maxCost[s][u];
            if (maxCost[s][v]<w) maxCost[s][v]=w;
            dfsCost(s,v,u);
        }
    }
public:
    void init(int n){
        this->n=n;
        memset(head,-1,sizeof(int)*(n+1));
        edge=0;
    }
    void addedge(int u,int v,int w){
        edges[edge].w=w,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
    }
    void findMaxCost(){
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++)
                maxCost[i][j]=0;
        for (int i=1;i<=n;i++){
            dfsCost(i,i,-1);
        }
    }
    int getMaxCost(int x,int y){
        return maxCost[x][y];
    }
}tree;

vector<Road>road;

int n,m;
int minTree;
bool can[maxn][maxn];

void initializer(){
    memset(can,0,sizeof(can));
    disjointSet.makeSet(n);
    road.clear();
    minTree=0;
    tree.init(n);
}

void input(){
    for (int i=0;i<m;i++){
        int a,b,w;
        scanf("%d%d%d",&a,&b,&w);
        road.push_back(Road(a,b,w));
    }
}

void Kruskal(){
    sort(road.begin(),road.end());
    for (vector<Road>::iterator it=road.begin();it!=road.end();it++){
        int u=it->u;
        int v=it->v;
        int w=it->w;
        if (disjointSet.findSet(u)!=disjointSet.findSet(v)){
            minTree+=w;
            disjointSet.unionSet(u,v);
            tree.addedge(u,v,w);
            tree.addedge(v,u,w);
            can[u][v]=can[v][u]=true;
        }
    }
    tree.findMaxCost();
}
void solve(){
    int ans=INF;
    for (vector<Road>::iterator it=road.begin();it!=road.end();it++){
        int u=it->u;
        int v=it->v;
        int w=it->w;
        if (!can[u][v]){
            ans=min(ans,minTree-tree.getMaxCost(u,v)+w);
        }
    }
    printf("%d %d
",minTree,ans);
}
int main()
{
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d%d",&n,&m);
        initializer();
        input();
        Kruskal();
        solve();
    }
    return 0;
}

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