HDU-4435-charge-station ( 2012 Asia Tianjin Regional Contest )

charge-station

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1025    Accepted Submission(s): 535

Problem Description

There are 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 has limited energy and can only travel by no more than D meters. Before it was run out of energy, it should be charged in some oil station. Under M^3's despotic power, the judge is forced to build several oil stations in some of the cities. The judge must build an oil station in city 1 and building other oil stations is up to his choice as long as M^3 can successfully travel around all the cities.
Building an oil station in city i will cost 2 ^i-1 MMMB. Please help the judge calculate out the minimum cost to build the oil stations in order to fulfill M^3's will.

 

Input

There are several test cases (no more than 50), each case begin with two integer N, D (the number of cities and the maximum distance the car can run after charged, 0 < N ≤ 128).
Then follows N lines and line i will contain two numbers x, y(0 ≤ x, y ≤ 1000), indicating the coordinate of city i.
The distance between city i and city j will 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, output the minimum cost to build the oil stations in the binary form without leading zeros.
If it's impossible to visit all the cities even after all oil stations are build, output -1 instead.

 

Sample Input

   
   
   
   

    
    
    
    3 3
0 0
0 3
0 1

3 2
0 0
0 3
0 1

3 1
0 0
0 3
0 1

16 23
30 40
37 52
49 49
52 64
31 62
52 33
42 41
52 41
57 58
62 42
42 57
27 68
43 67
58 48
58 27
37 69
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    11
111
-1
10111011

    
    
    
    
 
     
 
      Hint 
     
 In case 1, the judge should select (0, 0) and (0, 3) as the oil station which result in the visiting route: 1->3->2->3->1. And the cost is 2^(1-1) + 2^(2-1) = 3. 
    
    
    
    
     
   
   
   
   

 

Source

2012 Asia Tianjin Regional Contest

 

这题还比较复杂吧。。

思路：就是去寻找最小的消耗，有点类似贪心

贴个代码！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<queue>
#define inf 1<<30  
#define N 105
#define Min(a,b) ((a)<(b)?(a):(b))
#define Max(a,b) ((a)>(b)?(a):(b))
#define LL long long
using namespace std;

struct Point
{
    int x,y;
}p[N];
int n,d;
int path[N][N];
int ok[N];
double dist(int i,int j)
{
    return sqrt((double)(p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
}
bool bfs()
{
    bool vis[N];
    int dist[N];
    queue<int>que;
    memset(vis, 0, sizeof(vis));
    for(int i=0;i<n;i++)
    {
        if(ok[i]) dist[i]=0;
        else dist[i]=inf;
    }
    que.push(0);vis[0]=true;
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        for(int i=0;i<n;i++)
        {
            if(!vis[i]&&path[u][i]<=d)
            {
                dist[i]=min(dist[i],dist[u]+path[u][i]);
                if(ok[i])
                {
                    que.push(i);
                    vis[i]=true;
                }
            }
        }
    }
    for(int i=0;i<n;i++)
    {
        if(ok[i]&&!vis[i]) return false;
        if(!ok[i]&&dist[i]*2>d) return false;
    }
    return true;
}

void fun()
{
    for(int i=0;i<n;i++) ok[i]=1;
    if(!bfs()) {puts("-1");return; }
    for(int i=n-1;i>0;i--)
    {
        ok[i]=0;
        if(!bfs()) ok[i]=1;
    }
    int j=n-1;
    while(!ok[j]) j--;
    for(int i=j;i>=0;i--) printf("%d",ok[i]);
    putchar(10);
}

int main()
{
    while(scanf("%d%d",&n,&d)!=EOF)
    {
        for(int i=0;i<n;i++) scanf("%d%d",&p[i].x,&p[i].y);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                path[i][j]=ceil(dist(i,j));
            }
        }
        fun();
    }
    return 0;
}