UESTC 1636 梦后楼台高锁，酒醒帘幕低垂

题意：求一条路径，使得这条边连接1到n，求边权值的最大值与最小值的差

题解：最小生成树，对边权排序，可以枚举边的最大和最小的值，判断能否使得1和n连通

#include <bits/stdc++.h>
#define ll long long
#define maxn 1010
using namespace std;
struct edge{
    ll from,to,weight;
};
int cmp(edge a,edge b){
    return a.weight<b.weight;
}
vector<edge>edges;
int fa[maxn];
int find(int x){
    return x==fa[x]?x:(fa[x] = find(fa[x]));
}
void init(int n){
    for(int i=1;i<=n;i++) fa[i] = i;
}
int main(){
    ll n,m,i,j,a,b,c, ma = -1e17,mi = 1e17, ans = 1e18;
    cin>>n>>m;
    for(i=0;i<m;i++){
        cin>>a>>b>>c;
        edges.push_back((edge){a,b,c});
    }
    sort(edges.begin(), edges.end(), cmp);
    for(i=0;i<m;i++){
        init(n);
        mi = 1e17;
        ma = -1e18;
        for(j=i;j<m;j++){
            edge e = edges[j];
            mi = min(e.weight, mi);
            ma = max(e.weight, ma);
            int fau = find(e.from);
            int fav = find(e.to);
            if(fau != fav) fa[fau] = fav;
            if(find(n) == find(1)) break;
        }
        if(j<m) ans = min(ans, ma-mi);
    }
    cout<<ans<<endl;
    return 0;
}