timus 1982 Electrification Plan(最小生成树)

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

【分析】将任意两个带有发电站的城市之间的距离赋为0，然后prim求最小生成树。(这么简单的思路我居然没有想到，还是学长教我的)。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 10000000
#define mod 10000
typedef long long ll;
using namespace std;
const int N=6005;
const int M=50000;
int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
int a[N],w[N][N],vis[N],lowcost[N];
int n,m,k;
void prim()
{
    int sum=0;lowcost[1]=-1;
    for(int i=2;i<=n;i++){
        lowcost[i]=w[1][i];
    }
    for(int i=1;i<n;i++){
        int minn=inf,k;
        for(int j=1;j<=n;j++){
            if(lowcost[j]!=-1&&lowcost[j]<minn){
                k=j;minn=lowcost[j];
            }
        }
        sum+=minn;
        lowcost[k]=-1;
        for(int j=1;j<=n;j++){
            lowcost[j]=min(lowcost[j],w[k][j]);
        }
    }
    printf("%d
",sum);
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)scanf("%d",&w[i][j]);
    for(int i=1;i<=m;i++)for(int j=1;j<=m;j++)w[a[i]][a[j]]=0;
    prim();
    return 0;
}