53 lines
1.3 KiB
C++
53 lines
1.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 20010, M = 100010;
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int n, m, x, y;
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// p[i]表示i所属的集合，用p[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 p[N << 1]; //并查集数组
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int find(int x) {
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if (x == p[x]) return x;
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return p[x] = find(p[x]);
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}
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//合并集合
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bool join(int a, int b) {
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if (find(a) == find(b)) return false;
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p[find(a)] = find(b);
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return true;
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}
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//关系
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struct Node {
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int x, y, v; //冲突值
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//排序函数，按冲突值大小排序，大的在前
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const bool operator<(const Node &t) const {
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return v > t.v;
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}
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} a[M];
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int main() {
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//读入
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scanf("%d %d", &n, &m);
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for (int i = 1; i <= m; i++) scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].v);
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//排序,按冲突值由大到小排序
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sort(a + 1, a + m + 1);
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//初始化扩展域并查集，每个人都是自己的祖先
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for (int i = 1; i <= 2 * n; i++) p[i] = i;
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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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int px = find(a[i].x), py = find(a[i].y);
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if (px == py) {
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printf("%d\n", a[i].v);
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exit(0);
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} else
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join(a[i].y + n, px), join(a[i].x + n, py);
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}
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printf("%d\n", 0);
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return 0;
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} |