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 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 ×  ntable 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 
Problem Source: Open Ural FU Championship 2013

Tags:  graph theory   (

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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。

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<cstring>

 4 #define INF 0x3f3f3f3f

 5 using namespace std;

 6 

 7 int edge[110][110];

 8 int vis[110],dis[110];

 9 bool flag[110];

10 int n,k,ans;

11 

12 void prim()

13 {

14     int u=1,minw;

15     for(int i=1;i<=n;i++)

16     {

17         vis[i]=0;

18         dis[i]=edge[u][i];

19     }

20     vis[u]=1;

21     for(int i=1;i<n;i++)

22     {

23         minw=INF;

24         for(int j=1;j<=n;j++)

25         {

26             if(!vis[j] && dis[j]<minw)

27             {

28                 minw=dis[j];

29                 u=j;

30             }

31         }

32         ans+=minw;

33         vis[u]=1;

34         for(int j=1;j<=n;j++)

35         {

36             if(!vis[j] && edge[u][j]<dis[j])

37                 dis[j]=edge[u][j];

38         }

39     }

40 }

41 

42 int main()

43 {

44     int d;

45     while(scanf("%d%d",&n,&k)!=EOF)

46     {

47         memset(vis,0,sizeof(vis));

48         memset(flag,false,sizeof(flag));

49         ans=0;

50         for(int i=0;i<k;i++)

51         {

52             scanf("%d",&d);

53             flag[d]=true;

54         }

55         for(int i=1;i<=n;i++)

56         {

57             for(int j=1;j<=n;j++)

58             {

59                 scanf("%d",&edge[i][j]);

60             }

61         }

62         for(int i=1;i<=n;i++)

63         {

64             for(int j=1;j<=n;j++)

65             {

66                 if(flag[i]&&flag[j])

67                     edge[i][j]=0;

68             }

69         }

70         prim();

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

72     }

73     return 0;

74 }

View Code

 

 

AC代码2：

  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 }

View Code