58 lines
1.5 KiB
C++
58 lines
1.5 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 20010;
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const int M = 100010;
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int n, m, x, y;
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//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；
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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
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//要深入理解这个递归并压缩的过程
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int find(int x) {
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if (fa[x] != x) fa[x] = find(fa[x]);
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return fa[x];
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}
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//加入家族集合中
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void join(int c1, int c2) {
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int f1 = find(c1), f2 = find(c2);
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if (f1 != f2)fa[f1] = f2;
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}
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//关系
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struct node {
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int x; //第一个人
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int y; //第二个人
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int v; //冲突值
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} a[M];
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//排序函数，按冲突值大小排序，大的在前
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bool cmp(const node &a, const node &b) {
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return a.v > b.v;
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}
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int main() {
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//读入
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cin >> n >> m;
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for (int i = 1; i <= m; i++) cin >> a[i].x >> a[i].y >> a[i].v;
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//排序
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sort(a + 1, a + m + 1, cmp);
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//初始化并查集，每个人都是自己的祖先
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for (int i = 1; i <= 2 * n; i++)fa[i] = i;//注意这里是2*n
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//处理m组关系
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for (int i = 1; i <= m; i++) {
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x = find(a[i].x);
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y = find(a[i].y);
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if (x == y) {
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cout << a[i].v << endl;
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return 0;
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} else
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join(a[i].y + n, x), join(a[i].x + n, y);
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}
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cout << 0 << endl;
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return 0;
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}
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