给出N个点的坐标 边的权值为两点间的距离 删掉其中某点 求最小生成树的权值和 要求这权值最小
因为最多50个点 所以具体是删哪个点 用枚举
假如有4个点 就要求4次最小生成树 分别是2 3 4 | 1 3 4 | 1 2 4 | 1 2 3 这些点的
Sample Input
2
5
0 0
1 0
18 0
0 1
1 1
3
0 0
1 0
0 1
Sample Output
3.00
1.00
prim
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # define LL long long 7 using namespace std ; 8 9 const int INF=0x3f3f3f3f; 10 const int MAXN=110; 11 bool vis[MAXN]; 12 double lowc[MAXN]; 13 double cost[MAXN][MAXN] ; 14 15 struct poin 16 { 17 int x ; 18 int y ; 19 }p[110] , q[110]; 20 21 double Prim(int n) 22 { 23 int i , j ; 24 for (i = 0 ; i < n ; i++) 25 for (j = 0 ; j < n ; j++) 26 cost[i][j] = INF ; 27 for (i = 0 ; i < n ; i++) 28 for (j = i+1 ; j < n ; j++) 29 { 30 double t = sqrt((double)(q[i].x - q[j].x) * (q[i].x - q[j].x) + (q[i].y - q[j].y) * (q[i].y - q[j].y)) ; 31 cost[i][j] = t ; 32 cost[j][i] = t ; 33 } 34 double ans=0; 35 memset(vis,false,sizeof(vis)); 36 vis[0]=true; 37 for(int i=1;i<n;i++)lowc[i]=cost[0][i]; 38 for(int i=1;i<n;i++) 39 { 40 double minc=INF; 41 int p=-1; 42 for(int j=0;j<n;j++) 43 if(!vis[j]&&minc>lowc[j]) 44 { 45 minc=lowc[j]; 46 p=j; 47 } 48 if(minc==INF)return -1;//原图不连通 49 ans+=minc; 50 vis[p]=true; 51 for(int j=0;j<n;j++) 52 if(!vis[j]&&lowc[j]>cost[p][j]) 53 lowc[j]=cost[p][j]; 54 } 55 return ans; 56 } 57 58 int main() 59 { 60 61 // freopen("in.txt","r",stdin) ; 62 int T ; 63 scanf("%d" , &T) ; 64 while(T--) 65 { 66 int n ; 67 scanf("%d" , &n) ; 68 int i , j ; 69 double MIN = INF ; 70 for (i = 0 ; i < n ; i++) 71 scanf("%d %d" , &p[i].x , &p[i].y) ; 72 73 for (i = 0 ; i < n ; i++) 74 { 75 int k = 0 ; 76 for (j = 0 ; j < n ; j++) 77 { 78 if (i == j) 79 continue ; 80 q[k].x = p[j].x ; 81 q[k].y = p[j].y ; 82 k++ ; 83 } 84 double ans = Prim(n-1) ; 85 if (ans < MIN) 86 MIN = ans ; 87 } 88 printf("%.2lf " , MIN) ; 89 90 } 91 return 0 ; 92 }