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 }
