POJ 2349 Arctic Network（最小生成树）

Arctic Network

Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel. 
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts. 

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

Input

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

Output

For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

Sample Input

1
2 4
0 100
0 300
0 600
150 750

Sample Output

212.13

题目大意：DND组织想要通过无线网络把一些哨站连通。他们使用两种不同的科技，一种是卫星通信，另一种是无线电通信。当两个哨点同时拥有卫星通信技术时，他们可以无视距离远近地进行交流，而拥有无线电通信技术的两个哨点必须在距离D以内才能交流。D与无线电波发射器的能量大小有关，当然，能量越大，花的钱越多。现在给出拥有卫星通信技术的哨点数S和拥有无线电通信的哨点数P，求使得所有哨点连通的最短的D。

解题思路：求最小生成树中的第P - S大的权值，感觉用Kruskal算法比较好做一点，因为取边之前已经排好序了，而Prim算法还要再排一次序。

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#define EPS 1e-8
const int maxn = 505;
int par[maxn],rank[maxn],x[maxn],y[maxn];
int n,m,cnt;
struct edge
{
    int from;
    int to;
    double dis;
};
edge edges[maxn * maxn];
void init()
{
    for(int i = 0;i <= n;i++){
        par[i] = i;
    }
    memset(rank,0,sizeof(rank));
}

int find(int x)
{
    return par[x] == x ? x : par[x] = find(par[x]);
}

void unite(int x,int y)
{
    x = find(x);
    y = find(y);
    if(x == y)
        return ;
    if(rank[x] < rank[y]){
        par[x] = y;
    }
    else{
        par[y] = x;
        if(rank[x] == rank[y])
            rank[x]++;
    }
}

bool same(int x,int y)
{
    return find(x) == find(y);
}

bool cmp(edge a,edge b)
{
    return a.dis < b.dis;
}

double Kruskal()
{
    std::sort(edges,edges + cnt,cmp);
    init();
    double res = 0;
    int k = 0;
    for(int i = 0;i < cnt;i++){
        edge e = edges[i];
        if(!same(e.from,e.to)){
            unite(e.from,e.to);
            if(k == n - m)
                break;
            res = e.dis;
            k++;
        }
    }
    return res;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&m,&n);
        for(int i = 0;i < n;i++){
            scanf("%d %d",&x[i],&y[i]);
        }
        cnt = 0;
        for(int i = 0;i < n;i++){
            for(int j = i;j < n;j++){
                edges[cnt].from = i;
                edges[cnt].to = j;
                edges[cnt].dis = pow(x[i] - x[j],2.0) + pow(y[i] - y[j],2.0);
                cnt++;
            }
        }
        printf("%.2f\n",sqrt(Kruskal()));
    }
    return 0;
}