洛谷 P1111 修复公路

链接：https://www.luogu.org/problemnew/show/P1111

思路：简单的并查集，一开始想的是每合并一次就遍历一遍看看是否全部连通了，看了题解才知道只要用一个变量标志连通块数量即可，每次合并连通块数量减一

代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<vector>
#include<stack>
#include<string>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int M = int(1e5) * 2 + 5;
//vector<int> v;
struct node
{
    int x, y, t;
}a[M];
bool cmp(node a, node b)
{
    return a.t < b.t;
}

int root[M];
void init()
{
    for (int i = 0; i < M; i++) root[i] = i;
}
int Find(int x)
{
    int r = x;
    while (root[r] != r)
    {
        r = root[r];
    }
    int i;
    while (root[x] != r)
    {
        i = root[x];
        root[x] = r;
        x = i;
    }
    return r;
}
void merge(int x, int y)
{
    x = Find(x);
    y = Find(y);
    if (x != y) root[x] = y;
}
signed main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        cin >> a[i].x >> a[i].y >> a[i].t;
    }
    sort(a, a + m, cmp);
    init();
    int cnt = n;
    for (int i = 0; i < m; i++)
    {
        int x = Find(a[i].x);
        int y = Find(a[i].y);
        if (x != y)
        {
            cnt--;
            merge(x, y);
        }
        if (cnt == 1)
        {
            cout << a[i].t;
            return 0;
        }
    }
    cout << -1;
    return 0;
}

备注：每次合并前要检查两个村庄是否属于同一个连通块