ural 1982. Electrification Plan 【最小生成树】

1982. 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 incij 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 × ntable of integers {cij} (0 ≤ cij ≤ 105). It is guaranteed that cij = cji, cij > 0 for i ≠ j, cii = 0.

Output

Output the minimum cost to electrify all the cities.

Sample

inputoutput

4 21 40 2 4 32 0 5 24 5 0 13 2 1 0

3

Problem Author: Mikhail Rubinchik 

/*    题意：给你n个城市和k个发电站的位置,以及每个城市之间拉电线的费用,          求使所有城市都能得到供电的最小花费    类型：最小生成树    分析：把有发电站的城市之间的费用全部改成0,再跑一次最小生成树即可*/#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<iostream>using namespace std;#define int_inf ((1<<31)-1)#define ll_inf ((1ll<<63)-1)int lowcost[100000+100];int vis[100000+100];int Map[300][300];bool dian[100000+100];int n;int prime(){    for(int i=1;i<=n;i++){        lowcost[i]=Map[1][i];    }    int Min;    int ans=0;    lowcost[1]=0;    vis[1]=1;    for(int i=1;i<n;i++){        Min=int_inf;        int k;        for(int j=1;j<=n;j++){            if(!vis[j]&&Min>lowcost[j]){                Min=lowcost[j];                k=j;            }        }        vis[k]=true;        ans+=Min;        for(int j=1;j<=n;j++){            if(!vis[j]&&lowcost[j]>Map[k][j])                lowcost[j]=Map[k][j];        }    }    return ans;}int main(){    int m;    while(scanf("%d%d",&n,&m)!=EOF){        memset(dian,0,sizeof(dian));        memset(vis,0,sizeof(vis));        for(int i=1;i<=m;i++){            int a;            scanf("%d",&a);            dian[a]=1;        }        for(int i=1;i<=n;i++){            for(int j=1;j<=n;j++){                int c;                scanf("%d",&c);                Map[i][j]=c;                if(dian[i]==1&&dian[j]==1)Map[i][j]=0;            }        }        int result=prime();        printf("%d\n",result);    }}