1000 - Spoon Devil's 3-D Matrix
Time Limit:1s Memory Limit:32MByte
Submissions:208Solved:65
DESCRIPTION
Spoon Devil build a 3-D matrix, and he(or she) wants to know if he builds some bases what's the shortest distance to connect all of them.
INPUT
There are multiple test cases. The first line of input contains an integer
T
-th point.
OUTPUT
For each test case, output a line, which should accurately rounded to two decimals.
SAMPLE INPUT
221 1 02 2 031 2 30 0 01 1 1
SAMPLE OUTPUT
1.413.97
SOLUTION
思路:
三维点的MST(Krusal)
#include<iostream> #include<cmath> #include<string> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; const int MAXN=55; const int MAXM=1200; int pre[MAXN]; struct node { double x,y,z; node() { x=y=z=0; } }Node[MAXN]; struct edge { int s,e; double d; edge() { s=e=d=0; } }Edge[MAXM]; bool cmp(edge a, edge b) { return a.d<b.d; } int father(int x) { if(pre[x]==x) return x; else { pre[x]=father(pre[x]); return pre[x]; } } double krusal(int n) { double cost=0; for(int i=0;i<MAXN;i++)pre[i]=i; int cnt=0; int index=0; while(cnt<n-1) { int ps=father(Edge[index].s); int pe=father(Edge[index].e); if(ps!=pe) { pre[ps]=pe; cost+=Edge[index].d; cnt++; } index++; } return cost; } int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%lf%lf%lf",&Node[i].x,&Node[i].y,&Node[i].z); } int cnt=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { Edge[cnt].s=i;Edge[cnt].e=j; Edge[cnt].d=sqrt(pow(fabs(Node[i].x-Node[j].x),2.0)+pow(fabs(Node[i].y-Node[j].y),2.0)+pow(fabs(Node[i].z-Node[j].z),2.0)); cnt++; } } sort(Edge,Edge+cnt,cmp); printf("%.2lf ",krusal(n)); } return 0; }