HDU 4435 charge-station () BFS graph theory problem

e-charge-stationTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i6 4u Submit Status Practice HDU 4435Appoint Description:

Description

There is n cities in M^3 ' s empire. M^3 owns a palace and a car and the palace resides in City 1. One day, she wants-to-travel around all the cities from her palace and finally back to her home. However, her car had limited energy and can only travel by no more than D meters. Before it is run out of the energy, it should is charged in some. Under M^3 ' s despotic power, the judge is forced to build several oil stations in some of the cities. The judge must build a oil station in City 1 and building other oil stations are up to his choice as long as m^3 can Succe Ssfully travel around all the cities.
Building An, in city I'll cost 2 i-1 MMMB. Please help the judge calculate out of the minimum cost to build the oil stations in order to fulfill m^3 ' s would.

Input

There is several test cases (no more than), each case begin with the integer N, D (the number of the cities and the Maximu M distance the car can run after charged, 0 < n≤128).
Then follows N lines and line I'll contain II numbers x, y (0≤x, y≤1000), indicating the coordinate of city I.
The distance between city I and City J would be Ceil (sqrt ((XI-XJ) 2 + (YI-YJ) 2)). (Ceil means rounding the number up, e.g. ceil (4.1) = 5)

Output

For each case, the output of the minimum cost to build of the oil stations in the binary form without leading zeros.
If it ' s impossible to visit all the cities even after all oil stations is build, output-1 instead.

Sample Input

3 30 00 30 13 20 00 30 13 10 00 30 116 2330 4037 5249 4952 6431 6252 3342 4152 4157 5862 4242 5727 6843 6758 4858 2737 69

Sample Output

11111-110111011

Hint

In case 1, the judge should select (0, 0) and (0, 3) as the which result in the visiting route:1->3->2- >3->1. And the cost is 2^ (1-1) + 2^ (2-1) = 3.
The proposal is to give the N and D, below the coordinates of n points, respectively, the car can be used to refuel a distance of D. But in the middle can refuel, ask where refueling can make the car to exercise a full round trip, and require the establishment of gas stations to spend the least amount,
Output is binary judgment, 1 for build, 0 for not established

1#include <stdio.h>2#include <iostream>3#include <string.h>4#include <queue>5#include <math.h>6#include <algorithm>7 using namespacestd;8 Const intmaxn= Max;9 Const intinf=999999999;Ten intMAP[MAXN][MAXN]; One intn,d; A DoubleX[MAXN],Y[MAXN]; - intDIS[MAXN]; - BOOLVIS[MAXN]; the  - intANS[MAXN]; -  - intCheck () { +memset (Vis,false,sizeof(Vis)); -        for(intI=1; i<=n;i++){ +           if(Ans[i]) Adis[i]=0; at           Else -dis[i]=inf; -       } -queue<int>que; -Que.push (1); -vis[1]=true; in      while(!Que.empty ()) { -        intu=Que.front (); to Que.pop (); +            for(intI=1; i<=n;i++){ -               if(!vis[i]&&map[u][i]<=d) { theDis[i]=min (dis[i],dis[u]+map[u][i]); *                    if(Ans[i]) { $vis[i]=true;Panax Notoginseng Que.push (i); -                    } the               } +           }           A     } the      for(intI=1; i<=n;i++){ +        if(ans[i]==1&&!Vis[i]) -            return false; $        Else if(ans[i]==0&&dis[i]*2>d) $            return false; -     } -   return true; the         - }Wuyi  the intMain () { -         while(SCANF ("%d%d", &n,&d)! =EOF) { Wumemset (Map,0,sizeof(map)); -               for(intI=1; i<=n;i++){ Aboutscanf"%LF%LF",&x[i],&y[i]); $              } -               for(intI=1; i<=n;i++){ -                  for(intj=1; j<=n;j++){ -Map[i][j]=ceil (sqrt ((X[I]-X[J)) * (X[i]-x[j]) + (Y[i]-y[j]) * (y[i]-Y[j] )); A                 } +              } the               for(intI=1; i<=n;i++){ -ans[i]=1; $              } the               if(!Check ()) { theprintf"-1\n"); the                 Continue;  the               } -  in                for(inti=n;i>=1; i--){ theans[i]=0; the                   if(!check ()) Aboutans[i]=1; the               } the             inti; the          for(i=n;i>=1; i--){ +             if(ans[i]==1) -                  Break; the         }Bayi          for(intj=i;j>=1; j--) theprintf"%d", Ans[j]); theprintf"\ n"); -        } -        return 0; the}

HDU 4435 charge-station () BFS graph theory problem