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。

  1 //STATUS:C++_AC_31MS_401KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //#pragma comment(linker,"/STACK:102400000,102400000")

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef __int64 LL;

 34 typedef unsigned __int64 ULL;

 35 //const

 36 const int N=110;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=95041567,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-8;

 41 const double OO=1e15;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 struct Edge{

 59     int u,v,val;

 60     bool operator < (const Edge& a)const {

 61         return val<a.val;

 62     }

 63 }e[N*N];

 64 int n,k;

 65 int id[N],p[N],w[N][N];

 66 

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

 68 

 69 int main()

 70 {

 71  //   freopen("in.txt","r",stdin);

 72     int i,j,a,x,y,ans,cnt;

 73     while(~scanf("%d%d",&n,&k))

 74     {

 75         mem(id,0);

 76         for(i=0;i<k;i++){

 77             scanf("%d",&a);

 78             id[a]=1;

 79         }

 80         k=2;

 81         for(i=1;i<=n;i++){

 82             if(id[i])continue;

 83             id[i]=k++;

 84         }

 85         mem(w,INF);

 86         for(i=1;i<=n;i++){

 87             for(j=1;j<=n;j++){

 88                 scanf("%d",&a);

 89                 w[id[i]][id[j]]=Min(w[id[i]][id[j]],a);

 90             }

 91         }

 92         cnt=0;

 93         for(i=1;i<k;i++){

 94             for(j=i+1;j<k;j++){

 95                 e[cnt].u=i,e[cnt].v=j;

 96                 e[cnt].val=w[i][j];

 97                 cnt++;

 98             }

 99         }

100         sort(e,e+cnt);

101         ans=0;

102         for(i=1;i<k;i++)p[i]=i;

103         for(i=0;i<cnt;i++){

104             x=find(e[i].u);y=find(e[i].v);

105             if(x!=y){

106                 p[y]=x;

107                 ans+=e[i].val;

108             }

109         }

110 

111         printf("%d\n",ans);

112     }

113     return 0;

114 }