题目大意：给你一些点，他们可以连通，如果距离超过了d那么就要经过加油站，建立加油站的费用为第i个点是2^(i-1)。求费用最小，输出二进制表示的最小费用。

费用和sum最坏等于=2^0+2^1+……+2^(n-1)。所以最高位为0这个数字才会最小，从最高位暴力枚举如果删掉这个点之后图是连通的那么就可以删掉，否则不可以。

求图是否连通的时候可以爆搜求解，也可以并查集。

charge-station

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <time.h>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
///#define LL long long
#define LL __int64
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007

using namespace std;

const int maxn = 210;

bool vis[maxn];
int mp[maxn][maxn];

struct node
{
    int x, y;
} f[maxn];

int n, d;

double Dis(int a, int b)
{
    return sqrt(((f[a].x-f[b].x)*(f[a].x-f[b].x) + (f[a].y-f[b].y)*(f[a].y-f[b].y))*1.0);
}

bool bfs()
{
    bool flag[maxn];
    int dis[maxn];
    memset(flag, false, sizeof(flag));
    for(int i = 1; i <= n; i++)
    {
        if(vis[i])
        {
            dis[i] = 0;
            continue;
        }
        dis[i] = INF;
    }
    queue<int> que;
    que.push(1);
    flag[1] = true;
    while(!que.empty())
    {
        int x = que.front();
        que.pop();
        for(int i = 1; i <= n; i++)
        {
            if(flag[i] || mp[x][i] > d) continue;
            dis[i] = min(dis[i], dis[x]+mp[x][i]);
            if(vis[i])
            {
                flag[i] = true;
                que.push(i);
            }
        }
    }
    for(int i = 1; i <= n; i++)
    {
        if(vis[i] && !flag[i]) return false;
        if(!vis[i] && 2*dis[i] > d) return false;
    }
    return true;
}

int main()
{
    while(~scanf("%d %d",&n, &d))
    {
        for(int i = 1; i <= n; i++) scanf("%d %d",&f[i].x, &f[i].y);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                mp[i][j] = ceil(Dis(i, j));
        for(int i = 1; i <= n; i++) vis[i] = true;
        if(!bfs())
        {
            cout<<-1<<endl;
            continue;
        }
        for(int i = n; i >= 1; i--)
        {
            vis[i] = false;
            if(!bfs()) vis[i] = true;
        }
        int x = n;
        while(!vis[x]) x--;
        for(int i = x; i >= 1; i--) cout<<vis[i];
        cout<<endl;
    }
    return 0;
}