1 /* 2 The first line of each test case contains 1 <= S <= 100, the number of satellite channels! 3 注意:S表示一共有多少个卫星,那么就是有 最多有S-1个通道! 然后将最小生成树中的后边的 S-1通道去掉就行了! 4 思路:最小生成树中的第 k 个最小边! 5 */ 6 //克鲁斯克尔算法..... 7 #include<iostream> 8 #include<cstdio> 9 #include<cstring> 10 #include<algorithm> 11 #include<cmath> 12 using namespace std; 13 14 double x[800], y[800]; 15 16 struct node{ 17 int u, v; 18 double d; 19 }; 20 21 bool cmp(node a, node b){ 22 return a.d < b.d; 23 } 24 25 int f[505]; 26 27 node nd[150000]; 28 double ret[505]; 29 30 int getFather(int x){ 31 return x==f[x] ? x : f[x]=getFather(f[x]); 32 } 33 34 bool Union(int a, int b){ 35 int fa=getFather(a), fb=getFather(b); 36 if(fa!=fb){ 37 f[fa]=fb; 38 return true; 39 } 40 return false; 41 } 42 43 int main(){ 44 int n, m; 45 int t; 46 scanf("%d", &t); 47 while(t--){ 48 scanf("%d%d", &m, &n); 49 for(int i=1; i<=n; ++i){ 50 scanf("%lf%lf", &x[i], &y[i]); 51 f[i]=i; 52 } 53 int cnt=0; 54 for(int i=1; i<n; ++i) 55 for(int j=i+1; j<=n; ++j){ 56 nd[cnt].u=i; 57 nd[cnt].v=j; 58 nd[cnt++].d=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j])); 59 } 60 sort(nd, nd+cnt, cmp); 61 int cc=0; 62 for(int i=0; i<cnt; ++i) 63 if(Union(nd[i].u, nd[i].v)) 64 ret[cc++]=nd[i].d; 65 for(int i=0; i<cc; ++i) 66 cout<<ret[i]<<"fdsf"<<endl; 67 printf("%.2lf\n", ret[n-m-1]); 68 } 69 return 0; 70 }
1 //prim算法....... 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 using namespace std; 8 const double INF = 0x3f3f3f3f*1.0; 9 double x[800], y[800]; 10 11 int n, m; 12 double map[505][505]; 13 int vis[505]; 14 15 double ret[505]; 16 17 void prim(){ 18 memset(vis, 0, sizeof(vis)); 19 vis[1]=1; 20 for(int i=2; i<=n; ++i) 21 ret[i]=INF; 22 int root=1, p; 23 for(int i=1; i<n; ++i){ 24 double minLen=INF; 25 for(int j=2; j<=n; ++j){ 26 if(!vis[j] && ret[j]>map[root][j]) 27 ret[j]=map[root][j]; 28 if(!vis[j] && minLen>ret[j]){ 29 minLen=ret[j]; 30 p=j; 31 } 32 } 33 root=p; 34 vis[root]=1; 35 } 36 } 37 38 int main(){ 39 40 int t; 41 scanf("%d", &t); 42 while(t--){ 43 scanf("%d%d", &m, &n); 44 for(int i=1; i<=n; ++i) 45 scanf("%lf%lf", &x[i], &y[i]); 46 for(int i=1; i<n; ++i) 47 for(int j=i+1; j<=n; ++j) 48 map[i][j]=map[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j])); 49 50 prim(); 51 sort(ret, ret+n+1); 52 53 printf("%.2lf\n", ret[n-m+1]); 54 } 55 return 0; 56 }
UvaOJ10369 - Arctic Network
时间: 2024-11-05 14:46:31