URAL-1982 Electrification Plan 最小生成树

　　题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1982

　　题意：无向图，给n个点，n^2条边，每条边有个一权值，其中有k个点有发电站，给出这k个点的编号，选择最小权值的边，求使得剩下的点都能接收到电。

　　发电站之间显然不能有边，那么把k个点合成一个点，然后在图上就MST就可以了。

//STATUS:C++_AC_31MS_401KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=110;
const int INF=0x3f3f3f3f;
const int MOD=95041567,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v,val;
    bool operator < (const Edge& a)const {
        return val<a.val;
    }
}e[N*N];
int n,k;
int id[N],p[N],w[N][N];

int find(int x){return p[x]==x?x:p[x]=find(p[x]);}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,a,x,y,ans,cnt;
    while(~scanf("%d%d",&n,&k))
    {
        mem(id,0);
        for(i=0;i<k;i++){
            scanf("%d",&a);
            id[a]=1;
        }
        k=2;
        for(i=1;i<=n;i++){
            if(id[i])continue;
            id[i]=k++;
        }
        mem(w,INF);
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&a);
                w[id[i]][id[j]]=Min(w[id[i]][id[j]],a);
            }
        }
        cnt=0;
        for(i=1;i<k;i++){
            for(j=i+1;j<k;j++){
                e[cnt].u=i,e[cnt].v=j;
                e[cnt].val=w[i][j];
                cnt++;
            }
        }
        sort(e,e+cnt);
        ans=0;
        for(i=1;i<k;i++)p[i]=i;
        for(i=0;i<cnt;i++){
            x=find(e[i].u);y=find(e[i].v);
            if(x!=y){
                p[y]=x;
                ans+=e[i].val;
            }
        }

        printf("%d
",ans);
    }
    return 0;
}