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。
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 }
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 }