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
,p
,w

;

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\n",ans);
}
return 0;
}