标签:clu ifd scan 题意 second station while vector testcase
http://poj.org/problem?id=2031
给出三维坐标系下的n个球体,求把它们联通的最小代价。
最小生成树加上一点计算几何。建图,若两球体原本有接触,则边权为0;否则边权为它们球心的距离-两者半径之和。这样来跑Prim就ok了。注意精度。
#include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #include<cstdio> #include<algorithm> #include<map> #include<set> #define rep(i,e) for(int i=0;i<(e);i++) #define rep1(i,e) for(int i=1;i<=(e);i++) #define repx(i,x,e) for(int i=(x);i<=(e);i++) #define X first #define Y second #define PB push_back #define MP make_pair #define mset(var,val) memset(var,val,sizeof(var)) #define scd(a) scanf("%d",&a) #define scdd(a,b) scanf("%d%d",&a,&b) #define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c) #define pd(a) printf("%d\n",a) #define scl(a) scanf("%lld",&a) #define scll(a,b) scanf("%lld%lld",&a,&b) #define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c) #define IOS ios::sync_with_stdio(false);cin.tie(0) #define lc idx<<1 #define rc idx<<1|1 #define rson mid+1,r,rc #define lson l,mid,lc using namespace std; typedef long long ll; template <class T> void test(T a) { cout<<a<<endl; } template <class T,class T2> void test(T a,T2 b) { cout<<a<<" "<<b<<endl; } template <class T,class T2,class T3> void test(T a,T2 b,T3 c) { cout<<a<<" "<<b<<" "<<c<<endl; } const int inf = 0x3f3f3f3f; const ll INF = 0x3f3f3f3f3f3f3f3fll; const ll mod = 1e9+7; int T; void testcase() { printf("Case %d: ",++T); } const int MAXN = 1e5+10; const int MAXM = 30; const double PI = acos(-1.0); const double eps = 1e-7; struct node{ double x,y,z,r; }p[110]; double dist(node a,node b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z)); } double g[110][110],lowc[110]; bool vis[110]; int main() { #ifdef LOCAL freopen("data.in","r",stdin); #endif // LOCAL int n; while(~scd(n)&&n){ for(int i=0;i<n;i++){ scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z,&p[i].r); } double maxx=0.0; for(int i=0;i<n;i++){ g[i][i]=0.0; for(int j=i+1;j<n;j++){ if(dist(p[i],p[j])-(p[i].r+p[j].r)<=eps){ g[i][j]=g[j][i]=0.0; }else{ g[i][j]=g[j][i]=dist(p[i],p[j])-(p[i].r+p[j].r); maxx=max(maxx,g[i][j]); } } } mset(vis,false); double ans=0; vis[0]=true; for(int i=1;i<n;i++) lowc[i]=g[0][i]; for(int i=1;i<n;i++){ double minc = maxx; int p=-1; for(int j=0;j<n;j++){ if(!vis[j]&&minc>lowc[j]){ minc=lowc[j]; p=j; } } ans+=minc; vis[p]=true; for(int j=0;j<n;j++){ if(!vis[j]&&lowc[j]-g[p][j]>eps){ lowc[j]=g[p][j]; } } } printf("%.3f\n",ans); } return 0; }
POJ - 2031 Building a Space Station(计算几何+最小生成树)
标签:clu ifd scan 题意 second station while vector testcase
原文地址:https://www.cnblogs.com/fht-litost/p/9349776.html