复习--F - Earth Hour（最短路，连接1 2 3个点）

F - Earth Hour

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Earth Hour is an annual international event created by the WWF (World Wide Fund for Nature/World Wildlife Fund), held on the last Saturday of March, that asks households and businesses to turn off their non-essential lights and electrical appliances for one hour to raise awareness towards the need to take action on climate change. 
To respond to the event of this year, the manager of Hunan University campus decides to turn off some street lights at night. Each street light can be viewed as a point in a plane, which casts flash in a circular area with certain radius. 
What's more, if two illuminated circles share one intersection or a point, they can be regarded as connected. 
Now the manager wants to turn off as many lights as possible, guaranteeing that the illuminated area of the library, the study room and the dormitory are still connected(directly or indirectly). So, at least the lights in these three places will not be turned off. 

 

Input

The first line contains a single integer T, which tells you there are T cases followed. 
In each case: 
The first line is an integer N( 3<=N<=200 ), means there are N street lights at total. 
Then there are N lines: each line contain 3 integers, X,Y,R,( 1<=X,Y,R<=1000 ), means the light in position(X,Y) can illuminate a circle area with the radius of R. Note that the 1st of the N lines is corresponding to the library, the 2nd line is corresponding to the study room, and the 3rd line is corresponding to the dorm. 

 

Output

One case per line, output the maximal number of lights that can be turned off. 
Note that if none of the lights is turned off and the three places are still not connected. Just output -1. 

 

Sample Input

 351 1 11 4 14 1 12 2 13 3 171 1 14 1 12 4 11 3 13 1 13 3 14 3 161 1 15 1 15 5 13 1 25 3 23 3 1 

 

Sample Output

-121 

 

题目大意，有n个灯，编号为1到n，给出每个灯的坐标和灯光照的范围，要求，1号2号3号必须开着，并且连接在一起，求可以关闭的最多的灯是多少？

题目转化一下也就是要用灯将1和2和3连接在一起，并且要最少的灯将他们连接在一起。

将3个连接在一起总会有一个节点，这个点分别连接其他的三个点的最短，用1 2 3 分别求出对其它所有点的最短路，在枚举所有点为节点，求出最小的值，就是需要最小的值把1 2 3 连接在一起，用n减去 得到结果

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define INF 300using namespace std;struct node{    int x , y , r ;} p[300];int mm[300][300] , a[300] ;int dis[300] , vis[300] ;void dij(int n,int u){    int i , j , min1 , k ;    memset(vis,0,sizeof(vis));    for(i = 0 ; i <= n ; i++)        dis[i] = mm[u][i] ;    dis[u] = 0 ;    vis[u] = 1 ;    for(i = 1 ; i <= n ; i++)    {        min1 = INF ;        for(j = 1 ; j <= n ; j++)            if( !vis[j] && dis[j] < min1 )        {            min1 = dis[j] ;            k = j ;        }        if(min1 == INF)            break;        vis[k] = 1 ;        for(j = 1 ; j <= n ; j++)            if( !vis[j] && dis[j] > dis[k] + mm[k][j] )                dis[j] = dis[k] + mm[k][j] ;    }}int main(){    int t , i , j , n , m ;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(i = 1 ; i <= n ; i++)        {            scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].r);            for(j = 1 ; j <= i ; j++)            {                double l = sqrt( (p[i].x-p[j].x)*(p[i].x-p[j].x)*1.0 + (p[i].y-p[j].y)*(p[i].y-p[j].y)*1.0 );                if(l <= ( p[i].r + p[j].r ))                    mm[i][j] = mm[j][i] = 1 ;                else                    mm[i][j] = mm[j][i] = INF ;            }        }        memset(a,0,sizeof(a));        dij(n,1);        for(i = 1 ; i <= n ; i++)            a[i] += dis[i] ;        dij(n,2);        for(i = 1 ; i <= n ; i++)            a[i] += dis[i] ;        dij(n,3);        for(i = 1 ; i <= n ; i++)            a[i] = n - (a[i] + dis[i] + 1) ;        int max1 = -1 ;        for(i = 1 ; i <= n ; i++)            if( a[i] > max1 )                max1 = a[i] ;        if(max1 < 0 )            printf("-1\n");        else        printf("%d\n", max1);    }    return 0;}