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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 20010;
const int M = 100010;
int n, m, x, y;
//fa[i]表示i所属的集合用fa[i+n]表示i所属集合的补集实现的很巧妙可以当成一个使用并查集的巧妙应用
int fa[N << 1]; //并查集数组,有时也写成 N << 1 ,从左到右读作N左移1位就是 2*N
//要深入理解这个递归并压缩的过程
int find(int x) {
if (fa[x] != x) fa[x] = find(fa[x]);
return fa[x];
}
//加入家族集合中
void join(int c1, int c2) {
int f1 = find(c1), f2 = find(c2);
if (f1 != f2)fa[f1] = f2;
}
//关系
struct node {
int x; //第一个人
int y; //第二个人
int v; //冲突值
} a[M];
//排序函数,按冲突值大小排序,大的在前
bool cmp(const node &a, const node &b) {
return a.v > b.v;
}
int main() {
//读入
cin >> n >> m;
for (int i = 1; i <= m; i++) cin >> a[i].x >> a[i].y >> a[i].v;
//排序
sort(a + 1, a + m + 1, cmp);
//初始化并查集,每个人都是自己的祖先
for (int i = 1; i <= 2 * n; i++)fa[i] = i;//注意这里是2*n
//处理m组关系
for (int i = 1; i <= m; i++) {
x = find(a[i].x);
y = find(a[i].y);
if (x == y) {
cout << a[i].v << endl;
return 0;
} else
join(a[i].y + n, x), join(a[i].x + n, y);
}
cout << 0 << endl;
return 0;
}