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题目连接:
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 23729 | Accepted: 6631 |
Description
Input
Output
Sample Input
4 0 0 0 0 1 1 1 1 2 1 0 3 0
Sample Output
1.000
Output
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const double inf=1e18; const int N=15e4+10; const int maxn=1e3+10; const double eps=1e-5; int n; double cost[maxn][maxn],dis[maxn][maxn],x[maxn],y[maxn],z[maxn]; int vis[maxn]; double get_dis(int a,int b) { return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b])); } int check(double x) { mst(vis,0); double sum=0,lowcost[maxn]; vis[1]=1; For(i,1,n)lowcost[i]=cost[1][i]-x*dis[1][i]; For(i,2,n) { double temp=inf; int k=-1; For(j,2,n) { if(!vis[j]&&lowcost[j]<temp) { k=j; temp=lowcost[j]; } } if(k==-1)break; vis[k]=1; sum+=temp; For(j,2,n) { if(!vis[j]&&cost[k][j]-x*dis[k][j]<lowcost[j]) lowcost[j]=cost[k][j]-x*dis[k][j]; } } if(sum>=0)return 1; return 0; } int main() { while(1) { read(n); if(n==0)break; For(i,1,n) { scanf("%lf%lf%lf",&x[i],&y[i],&z[i]); } For(i,1,n) For(j,i+1,n) { dis[i][j]=dis[j][i]=get_dis(i,j); cost[i][j]=cost[j][i]=abs(z[i]-z[j]); } double l=0.0,r=100.0; while(r-l>=eps) { double mid=(l+r)/2; if(check(mid))l=mid; else r=mid; } printf("%.3f\n",r); } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5684346.html