TOJ Open Ural FU Championship 2013 B

B. Electrification Plan

Time limit: 0.5 second
Memory limit: 64 MB

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 i, j it is possible to build a power line between them in c_ij 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 {c_ij} (0 ≤ c_ij ≤ 10⁵). It is guaranteed that c_ij = c_ji, c_ij > 0 for i ≠ j; c_ii = 0.

Output

Output the minimum cost to electrify all the cities.

Sample

input	output
4 2 1 4 0 2 4 3 2 0 5 2 4 5 0 1 3 2 1 0	3

Problem Author: Mikhail Rubinchik

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define maxn 1001
#define ll long long
#define INF 0x7fffffff
//#define ll __int64
int a[maxn];
int b[maxn][maxn];
struct Edge{
    int from, to, cost;
};
struct Node{
    int u, cost;
    bool operator<(const Node& rhs)const{
        return cost>rhs.cost;
    }
};
priority_queue<Node> q;
vector<Edge>edges;
vector<int>G[maxn];
int vis[maxn];
int m, n;
int from, to, cost;
void AddEdge(int from, int to, int cost){
    edges.push_back((Edge){ from, to, cost });
    int m = edges.size();
    G[from].push_back(m - 1);
    edges.push_back((Edge){ to, from, cost });
    G[to].push_back(m);
}
void init()
{
    memset(vis, 0, sizeof vis);
    for (int i = 0; i<n; i++)G[i].clear();
    edges.clear();
    while (!q.empty())q.pop();
}
void Prim()
{
    int cost = 0, t = 0;
    Node x;
    from=0;
    x.u = from;//选择起点
    x.cost = 0;
    q.push(x);
    while (!q.empty() && t<=n)
    {
        x = q.top(); q.pop();
        if (vis[x.u])continue;
        vis[x.u] = 1;
        cost += x.cost;
        t++;
        for (int i = 0; i<G[x.u].size(); i++){
            Edge e = edges[G[x.u][i]];
            if (!vis[e.to])q.push((Node){ e.to, e.cost });
        }
    }
    printf("%d\n", cost);
}
int main(){
    while (~scanf("%d%d", &n, &m)){
        memset(a, 0, sizeof a);
        memset(b, 0, sizeof b);
        init();
        for(int i=0;i<=n;i++)
        for(int j=0;j<=n;j++)AddEdge(i,j,INF);
        for (int i = 1; i <= m; i++){
            scanf("%d", &a[i]);
            AddEdge(0, a[i], 0);
        }
        for (int i = 1; i <= n;i++)
        for (int j = 1; j <= n; j++){
            scanf("%d", &b[i][j]);
            AddEdge(i, j, b[i][j]);
        }
        Prim();
    }
    return 0;
}