POJ 2349 Arctic Network【最小生成树】

拼搏现实的明天。 2022-05-25 04:08 68阅读 0赞

## Arctic Network ##

Time Limit: 2000MS Memory Limit: 65536K  
Total Submissions: 25250 Accepted: 7770

#### 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

#### 解题分析： ####

算出每个点到另一个点的距离，做一次最小生成树，求出n-m-1条边中最大的边值。

AC代码：

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    
    #define N 1100
    
    struct node{
        int x, y;
        double sum;
    }s[N*N];
    
    int f[N];
    
    bool cmp(node a, node b)
    {
        if(a.sum < b.sum)
            return 1;
        return 0;
    }
    
    double ff(int a, int b, int c, int d)
    {
        return sqrt(1.0*(a-b)*(a-b)+1.0*(c-d)*(c-d));
    }
    
    int Find(int v)
    {
        if(f[v] == v) return v;
        f[v] = Find(f[v]);
        return f[v];
    }
    
    int Union(int r1, int r2)
    {
        int t1 = Find(r1), t2 = Find(r2);
        if(t1 != t2){
            f[t1] = t2;
            return 1;
        }
        return 0;
    }
    
    double Kruskal(int n, int m, int len)
    {
        int i, k = 0;
        double Max;
        sort(s, s+len, cmp);
        for(i = 0; i < 1100; i++) f[i] = i;
        for(i = 0; i < len && k < n-m; i++){
            if(Union(s[i].x, s[i].y)){
                k++;
                Max = s[i].sum;
            }
        }
        return Max;
    }
    
    int main()
    {
        int x[N], y[N];
        int i, j, k, n, m, T;
        scanf("%d", &T);
        while(T--){
            scanf("%d%d", &m, &n);
            for(i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
            k = 0;
            for(i = 1; i < n; i++){
                for(j = i+1; j <= n; j++){
                    s[k].x = i;
                    s[k].y = j;
                    s[k].sum = ff(x[i], x[j], y[i], y[j]);
                    k++;
                }
            }
            printf("%.2f\n", Kruskal(n, m, k));//输出这只能用%f，不能用%lf，不然会错。
        }
    }