2728

题目大意：空间中有n个点，要把这n个点建成一棵树。树边有两个权值a、b，即对于点(x1,y1,z1)、(x2,y2,z2)间的边，a=abs(z1-z2)，b=sqrt((x1-x2)^2+(y1-y2)^2)。要求(∑a)/(∑b)的最小值，保留三位小数。答案取值范围(0,1e7)

观察数据范围，发现恰好还是可以二分答案的。

记此时的答案为k，则答案符合题意，当且仅当存在这样一棵树，使得(∑a)/(∑b)≤k，即∑(a-b*k)≤0

然后发现是个最小生成树问题。由于是完全图，需要用prim算法。

实数二分有毒啊。。。这里放两个代码。一个是我写的，还有一个实数二分是复制过来的。

#include<cstdio>#include<cstring>#include<iostream>#include<vector>#include<algorithm>#include<set>#include<cmath>#include<queue>using namespace std;#define rep(i,j,k) for(i=j;i<=k;++i)#define per(i,j,k) for(i=j;i>=k;--i)#define sqr(x) ((x)*(x))#define G getchar()#define LL long long#define pii pair<int,int>#define mkp make_pair#define X first#define Y second#define N 1005#define inf 10000001#define eps 0.000001int n,a[N],b[N],c[N],A[N][N];double B[N][N];double w[N][N],ans,Min[N];bool vis[N];void read(int &x){char ch=G;while(ch<48||ch>57)ch=G;for(x=0;ch>47&&ch<58;ch=G)x=x*10+ch-48;}void prim(){int i,j,x,y;double z;rep(i,2,n)Min[i]=inf,vis[i]=false;vis[x=1]=true;ans=0;rep(i,2,n){z=inf;rep(j,1,n)if(!vis[j])if((Min[j]=min(Min[j],w[x][j]))<z)z=Min[y=j];vis[x=y]=true;ans+=Min[y];}}bool check(double k){int i,j;rep(i,1,n)rep(j,1,n)if(i!=j)w[i][j]=A[i][j]-k*B[i][j];prim();return ans<=eps;}int main(){int i,j,tmp;LL l,r,mid;while(1){read(n);if(!n)break;tmp=0;rep(i,1,n){read(a[i]);read(b[i]);read(c[i]);tmp=max(tmp,c[i]);}rep(i,1,n)rep(j,1,n)A[i][j]=abs(c[i]-c[j]),B[i][j]=sqrt(sqr(a[i]-a[j])+sqr(b[i]-b[j]));for(l=0,r=1000000LL*tmp;l<=r;){mid=l+r>>1;if(check(mid/1000000.0))r=mid-1;else l=mid+1;}printf("%.3f\n",l/1000000.0);}return 0;}

#include<cstdio>#include<cstring>#include<iostream>#include<vector>#include<algorithm>#include<set>#include<cmath>#include<queue>using namespace std;#define rep(i,j,k) for(i=j;i<=k;++i)#define per(i,j,k) for(i=j;i>=k;--i)#define sqr(x) ((x)*(x))#define G getchar()#define LL long long#define pii pair<int,int>#define mkp make_pair#define X first#define Y second#define N 1005#define inf 10000001#define eps 0.000001int n,a[N],b[N],c[N],A[N][N];double B[N][N];double w[N][N],ans,Min[N];bool vis[N];void read(int &x){char ch=G;while(ch<48||ch>57)ch=G;for(x=0;ch>47&&ch<58;ch=G)x=x*10+ch-48;}void prim(){int i,j,x,y;double z;rep(i,2,n)Min[i]=inf,vis[i]=false;vis[x=1]=true;ans=0;rep(i,2,n){z=inf;rep(j,1,n)if(!vis[j])if((Min[j]=min(Min[j],w[x][j]))<z)z=Min[y=j];vis[x=y]=true;ans+=Min[y];}}bool check(double k){int i,j;rep(i,1,n)rep(j,1,n)if(i!=j)w[i][j]=A[i][j]-k*B[i][j];if(k==0.454)k=0.454;if(k==0.453)k=0.453;prim();return ans<=eps;}int main(){int i,j,tmp;while(1){read(n);if(!n)break;tmp=0;rep(i,1,n){read(a[i]);read(b[i]);read(c[i]);tmp=max(tmp,c[i]);}rep(i,1,n)rep(j,1,n)A[i][j]=abs(c[i]-c[j]),B[i][j]=sqrt(sqr(a[i]-a[j])+sqr(b[i]-b[j]));double l=0.0,r=100.0;            while(r-l>=eps)            {                double mid=(l+r)/2;                if(check(mid))r=mid;                else l=mid;            }            printf("%.3f\n",r);}return 0;}