Timusoj 1982. Electrification Plan

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

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

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 
Problem Source: Open Ural FU Championship 2013

Tags: graph theory  (hide tags for unsolved problems)

Difficulty: 144    Printable version    Submit solution    Discussion (4)
All submissions (2430)    All accepted submissions (894)    Solutions rating (630)

分析：

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。

AC代码1：

　　　　1、edge[i][j]=0是个很巧妙的设置。

2、求最小生成树，由于生成树是图的极小联通子图。最小生成树一定要包含图中所有的点。

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #define INF 0x3f3f3f3f
 using namespace std;

 int edge[][];
 int vis[],dis[];
 bool flag[];
 int n,k,ans;

 void prim()
 {
     int u=,minw;
     for(int i=;i<=n;i++)
     {
         vis[i]=;
         dis[i]=edge[u][i];
     }
     vis[u]=;
     for(int i=;i<n;i++)
     {
         minw=INF;
         for(int j=;j<=n;j++)
         {
             if(!vis[j] && dis[j]<minw)
             {
                 minw=dis[j];
                 u=j;
             }
         }
         ans+=minw;
         vis[u]=;
         for(int j=;j<=n;j++)
         {
             if(!vis[j] && edge[u][j]<dis[j])
                 dis[j]=edge[u][j];
         }
     }
 }

 int main()
 {
     int d;
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         memset(vis,,sizeof(vis));
         memset(flag,false,sizeof(flag));
         ans=;
         for(int i=;i<k;i++)
         {
             scanf("%d",&d);
             flag[d]=true;
         }
         for(int i=;i<=n;i++)
         {
             for(int j=;j<=n;j++)
             {
                 scanf("%d",&edge[i][j]);
             }
         }
         for(int i=;i<=n;i++)
         {
             for(int j=;j<=n;j++)
             {
                 if(flag[i]&&flag[j])
                     edge[i][j]=;
             }
         }
         prim();
         printf("%d\n",ans);
     }
     return ;
 }

AC代码2：

 //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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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,);
         for(i=;i<k;i++){
             scanf("%d",&a);
             id[a]=;
         }
         k=;
         for(i=;i<=n;i++){
             if(id[i])continue;
             id[i]=k++;
         }
         mem(w,INF);
         for(i=;i<=n;i++){
             for(j=;j<=n;j++){
                 scanf("%d",&a);
                 w[id[i]][id[j]]=Min(w[id[i]][id[j]],a);
             }
         }
         cnt=;
         for(i=;i<k;i++){
             for(j=i+;j<k;j++){
                 e[cnt].u=i,e[cnt].v=j;
                 e[cnt].val=w[i][j];
                 cnt++;
             }
         }
         sort(e,e+cnt);
         ans=;
         for(i=;i<k;i++)p[i]=i;
         for(i=;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 ;
 }