题目链接：
https://vjudge.net/problem/POJ-2485
题目大意：
求最小生成树中的最大边
思路：
是稠密图，用prim更好，但是规模不大，kruskal也可以过
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<sstream> 10 using namespace std; 11 typedef long long ll; 12 const int maxn = 3e5 + 10; 13 const int INF = 1 << 30; 14 int dir[4][2] = {1,0,0,1,-1,0,0,-1}; 15 int T, n, m, x; 16 struct edge 17 { 18 int u, v, w; 19 bool operator <(const edge& a)const 20 { 21 return w < a.w; 22 } 23 }; 24 edge a[maxn]; 25 int par[600], high[600]; 26 //初始化n个元素 27 void init(int n) 28 { 29 for(int i = 0; i < n; i++) 30 { 31 par[i] = i; 32 high[i] = 0; 33 } 34 } 35 //查询树的根 36 int Find(int x) 37 { 38 return par[x] == x ? x : par[x] = Find(par[x]);//路径压缩 39 } 40 void unite(int x, int y) 41 { 42 x = Find(x); 43 y = Find(y); 44 if(x == y)return; 45 if(high[x] < high[y])par[x] = y;//y的高度高，将x的父节点设置成y 46 else 47 { 48 par[y] = x; 49 if(high[x] == high[y])high[x]++; 50 } 51 } 52 bool same(int x, int y) 53 { 54 return Find(x) == Find(y); 55 } 56 void kruskal(int n, int m)//点数n，边数m 57 { 58 int ans = 0;//mst权值 59 int num= 0;//已经选择的边的边数 60 sort(a, a + m);//边进行排序 61 init(n);//初始化并查集 62 for(int i = 0; i < m; i++) 63 { 64 int u = a[i].u; 65 int v = a[i].v; 66 if(Find(u - 1) != Find(v - 1))//图最开始的下标是1，并查集是0 67 { 68 //printf("%d %d %d ", u, v, a[i].w); 69 //sum_mst += a[i].w; 70 ans = max(ans, a[i].w); 71 num++; 72 unite(u - 1, v - 1); 73 } 74 if(num >= n - 1)break; 75 } 76 //printf("weight of mst is %d ", sum_mst); 77 printf("%d ", ans); 78 } 79 int main() 80 { 81 cin >> T; 82 while(T--) 83 { 84 scanf("%d", &n); 85 int x; 86 int tot = 0; 87 for(int i = 1; i <= n; i++) 88 { 89 for(int j = 1; j <= n; j++) 90 { 91 scanf("%d", &x); 92 if(i == j)continue; 93 a[tot].u = i; 94 a[tot].v = j; 95 a[tot++].w = x; 96 } 97 } 98 kruskal(n, tot); 99 } 100 return 0; 101 }