• Electrification Plan MST


    Electrification Plan
    Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    Description

    Some country has n cities. The government has decided to electrify all these cities. At first, power stations in k different cities were built. The other cities should be connected with the power stations via power lines. For any cities ij it is possible to build a power line between them in cij roubles. The country is in crisis after a civil war, so the government decided to build only a few power lines. Of course from every city there must be a path along the lines to some city with a power station. Find the minimum possible cost to build all necessary power lines.

    Input

    The first line contains integers n and k (1 ≤ k ≤ n ≤ 100). The second line contains k different integers that are the numbers of the cities with power stations. The next n lines contain an n × n table of integers { cij} (0 ≤ cij ≤ 10 5). It is guaranteed that cij = cjicij > 0 for i ≠ j,cii = 0.

    Output

    Output the minimum cost to electrify all the cities.

    Sample Input

    inputoutput
    4 2
    1 4
    0 2 4 3
    2 0 5 2
    4 5 0 1
    3 2 1 0
    
    3
    int n,m;
    struct Edge
    {
        int from,to,cost;
    };
    
    int p[maxn];
    int find(int x)
    {
        if(x == p[x])
            return x;
        else
        {
            p[x] = find(p[x]);
            return p[x];
        }
    }
    
    vector<Edge> edges; 
    bool cmp(Edge e1,Edge e2)
    {
        return e1.cost < e2.cost;
    }
    
    
    int kruskal()
    {
        int ans = 0;
        rep(i,0,edges.size())
        {
            int x = find(edges[i].from);
            int y = find(edges[i].to);
            if(x != y)
            {
                p[x] = y;
                ans += edges[i].cost;
            }
        }
        return ans;
    }
    int main() 
    {
        //freopen("in.txt","r",stdin);
        while(cin>>n>>m)
        {
            int b;
            cin>>b;
            repf(i,1,n)
                p[i] = i;
            
            repf(i,2,m)
            {
                int a;
                scanf("%d",&a);
                p[a] = b;
            }
            edges.clear();
            repf(i,1,n)
                repf(j,1,n)
                {
                    int a;
                    scanf("%d",&a);
                    if(i < j)
                        edges.push_back((Edge){i,j,a});
                }
        
            sort(edges.begin(),edges.end(),cmp);
            
            cout<<kruskal()<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/DreamHighWithMe/p/3425418.html
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