题目链接：
https://vjudge.net/problem/POJ-2349
题目大意：
要在n个节点之间建立通信网络，其中m个节点可以用卫星直接连接，剩下的节点都要用线路连接，求剩下这些线路中最大的长度需要多长
思路：
还是MST的裸题，由于有m个节点可以用卫星连接，所以求出MST后，最长的m-1条边都可以用卫星连接，只需要求出第m长的边即可，直接用prim算法，保存mst每条边的长度，排序后直接输出第m长的边。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<queue> 7 #include<stack> 8 #include<map> 9 #include<set>10 #include<sstream>11 using namespace std;12 typedef long long ll;13 typedef pair<int, int> Pair;14 const int maxn = 1e3 + 10;15 const int INF = 1 << 30;16 int dir[4][2] = {1,0,0,1,-1,0,0,-1};17 int T, n, m;18 double Map[maxn][maxn];//存图19 double lowcost[maxn], ans[maxn];20 int mst[maxn], tot;21 Pair a[maxn];22 void prim(int u)//最小生成树起点23 {24 ????for(int i = 1; i <= n; i++)//初始化两个数组25 ????{26 ????????lowcost[i] = Map[u][i];27 ????????mst[i] = u;28 ????}29 ????mst[u] = -1;//设置成-1表示已经加入mst30 ????for(int i = 1; i <= n; i++)31 ????{32 ????????double minn = INF;33 ????????int v = -1;34 ????????//在lowcost数组中寻找未加入mst的最小值35 ????????for(int j = 1; j <= n; j++)36 ????????{37 ????????????if(mst[j] != -1 && lowcost[j] < minn)38 ????????????{39 ????????????????v = j;40 ????????????????minn = lowcost[j];41 ????????????}42 ????????}43 ????????if(v != -1)//v=-1表示未找到最小的边,44 ????????{//v表示当前距离mst最短的点45 ????????????//printf("%d %d %d\n", mst[v], v, lowcost[v]);//输出路径46 ????????????ans[tot++] = lowcost[v];47 ????????????mst[v] = -1;48 ????????????for(int j = 1; j <= n; j++)//更新最短边49 ????????????{50 ????????????????if(mst[j] != -1 && lowcost[j] > Map[v][j])51 ????????????????{52 ????????????????????lowcost[j] = Map[v][j];53 ????????????????????mst[j] = v;54 ????????????????}55 ????????????}56 ????????}57 ????}58 ????//printf("weight of mst is %d\n", sum_mst);59 ????sort(ans, ans + tot);60 ????printf("%.2f\n", ans[n - m - 1]);//输出第m长的边61 }62 int main()63 {64 ????cin >> T;65 ????while(T--)66 ????{67 ????????cin >> m >> n;68 ????????for(int i = 1; i <= n; i++)cin >> a[i].first >> a[i].second;69 ????????for(int i = 1; i <= n; i++)70 ????????{71 ????????????for(int j = 1; j <= n; j++)72 ????????????????Map[i][j] = Map[j][i] = sqrt((a[i].first - a[j].first) * (a[i].first - a[j].first) + (a[i].second - a[j].second) * (a[i].second - a[j].second));73 ????????}74 ????????tot = 0;75 ????????prim(1);76 ????}77 ????return 0;78 }POJ-2349 Arctic Network---MST的第m长的边
原文地址：https://www.cnblogs.com/fzl194/p/8727421.html