题目求解的是最小生成树中的最大边：
一次AC代码：
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;class data{public:int from, to, cost;};data Edge[10005];int par[2005];int N, M, num;int cmp(const data& a, const data& b) //从小到大排序{return a.cost < b.cost;}int finds(int x)//并查集查找函数{if(x == par[x])return x;elsereturn par[x] = finds(par[x]);}void join(int x, int y)//并查集合并函数{x = finds(x);y = finds(y);if(x != y)par[y] = x;}int Kruskal()//用Kruskal算法求解最大费用{int i, ans = 0;for(i = 0; i <= N; i++)par[i] = i;sort(Edge, Edge + num, cmp);//排序for(i = 0; i < num; i++){data e = Edge[i];if(finds(e.from) != finds(e.to)) //判断是否属于一个并查集{join(e.from, e.to);if(ans < e.cost)ans = e.cost;}}return ans;}int main(){int a, b, c;while(scanf("%d%d", &N, &M) != EOF){num = 0;for(int i = 0; i < M; i++){scanf("%d%d%d", &a, &b, &c);Edge[num].from = a;Edge[num].to = b;Edge[num++].cost = c;}int ans = Kruskal();printf("%d\n", ans);}return 0;}
，而更像是听见了天地间冥冥中的呼唤，
