$Kruskal$算法（334ms）$O(mlogm)$

核心：按边长从小到大枚举边，第一次使得1和n连通的边的长度即为答案

#include <algorithm>
#include <cstring>
#include <iostream>

using namespace std;

const int N = 100010, M = 400010;

struct Edge
{
    int a, b, c;
    bool operator<(const Edge &t) const
    {
        return c < t.c;
    }
} e[M];

int n, m;
int p[N];

int find(int x)
{
    if (p[x] != x)
        p[x] = find(p[x]);
    return p[x];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        e[i] = {a, b, c};
    }
    sort(e, e + m);
    for (int i = 1; i <= n; i++)
        p[i] = i;
    for (int i = 0; i < m; i++)
    {
        int a = e[i].a, b = e[i].b;
        a = find(a), b = find(b);
        if (a != b)
        {
            p[a] = b;
            if (find(1) == find(n))
            {
                printf("%d", e[i].c);
                break;
            }
        }
    }
    return 0;
}

堆优化$dijkstra$算法的变种用法（834ms）$O(mlogn)$

核心：$dist[j] = max(dist[t.y], w[i])$;

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>

#define x first
#define y second

using namespace std;

const int N = 100010, M = 400010;

typedef pair<int, int> PII;

int n, m;
int h[N], e[M], ne[M], w[M], idx;
int dist[N];
bool st[N];

void add(int a,int b,int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    priority_queue<PII, vector<PII>, greater<PII>> q;
    q.push({0, 1});
    dist[1] = 0;
    while (q.size())
    {
        auto t = q.top();
        q.pop();
        if(st[t.y])
            continue;
        st[t.y] = true;
        for (int i = h[t.y]; ~i;i=ne[i])
        {
            int j = e[i];
            if(dist[j]>max(dist[t.y],w[i]))
            {
                dist[j] = max(dist[t.y], w[i]);
                if(!st[j])
                    q.push({dist[j], j});
            }
        }
    }
    return dist[n];
}

int main()
{
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);
    for (int i = 0; i < m;i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c), add(b, a, c);
    }
    cout << dijkstra() << endl;
    return 0;
}