http://acm.hdu.edu.cn/showproblem.php?pid=3832

大致题意：给出n个灯的位置以及每个灯照亮的范围。问最后至多可以熄灭多少盏灯使得编号为1,2,3的点连通。

思路：若任意两灯的照亮范围相交，那么两点之间建边，权值为1。最后枚举一点使它到1,2,3点的距离最近，那么该距离之和+1（还有该点本身）代表必须不能熄灭的灯，那么熄灭的灯数就求出来了。

floyd会TLE。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#define LL long long
#define _LL __int64
using namespace std;
const int INF = 1<<28;
const int maxn = 210;

struct node
{
	int x,y;
	int r;
}p[maxn];
int n;
int Map[maxn][maxn];
int dis[3][maxn];
int vis[maxn];

void init()
{
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= n; j++)
		{
			if(i == j) Map[i][j] = 0;
			else Map[i][j] = INF;
		}
	}

	for(int i = 1; i < n; i++)
	{
		for(int j = i+1; j <= n; j++)
		{
			int dis = (p[i].x-p[j].x)*(p[i].x-p[j].x) + (p[i].y-p[j].y)*(p[i].y-p[j].y);
			if( dis <= (p[i].r+p[j].r)*(p[i].r+p[j].r) )
			{
				Map[i][j] = Map[j][i] = 1;
			}
		}
	}
}

void spfa(int s,int f)
{
	for(int i = 1; i <= n; i++)
		dis[f][i] = INF;

	queue<int> que;
	memset(vis,0,sizeof(vis));

	dis[f][s] = 0;
	vis[s] = 1;
	que.push(s);

	while(!que.empty())
	{
		int u = que.front();
		que.pop();
		vis[u] = 0;

		for(int i = 1; i <= n; i++)
		{
			if(Map[u][i] < INF && dis[f][i] > dis[f][u] + Map[u][i])
			{
				dis[f][i] = dis[f][u] + Map[u][i];
				if(!vis[i])
				{
					vis[i] = 1;
					que.push(i);
				}
			}
		}
	}
}

int main()
{
	int test;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d",&n);
		for(int i = 1; i <= n; i++)
			scanf("%d %d %d",&p[i].x,&p[i].y,&p[i].r);
		init();

		spfa(1,0);
		spfa(2,1);
		spfa(3,2);
		int ans = INF;
		for(int i = 1; i <= n; i++)
		{
			if(dis[0][i] + dis[1][i] + dis[2][i] < ans)
				ans = dis[0][i] + dis[1][i] + dis[2][i];
		}
		if(ans == INF)
			printf("-1\n");
		else printf("%d\n",n-ans-1);

	}
	return 0;
}