gcd最大生成树模板

出处：

ACM International Collegiate Programming Contest, Egyptian Collegiate Programming Contest
Arab Academy for Science, Technology and Maritime Transport, 2017

 

想法题：n=1e5. 有n*n/2条边，不能kruskal。

但是考虑一下，边权都是gcd，而gcd（x,y)<min(x,y),所以权值不同的数只有1e5个。所以依然用kruskal的想法，枚举所有不同权值的边，然后暴力枚举gcd为该边的两个数，将他们连起来。具体做法就是枚举该边的所有倍数

#include <iostream>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include<cstdio>
#include<vector>
#include<ctime>

#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
const double PI = acos(-1.0);
int a[maxn], vis[maxn], f[maxn];
int find(int x) { if (f[x] == x) { return x; }return f[x] = find(f[x]); }

int main()
{
    freopen("dream.in", "r", stdin);
    int t;
    cin >> t;
    int n;
    int x;
    rep(k, 1, t) {
        
        mmm(vis, 0);
        cin >> n;
        ll ans = 0;
        int tot = 0;
        int mx = 0;
        rep(i, 1, n)
        {
            scanf("%d", &x);
            if (vis[x]) { ans += x; continue; }
            vis[x] = 1;
            f[x] = x;
            a[++tot] = x;
            mx = max(mx, x);
        }
        int p = 0;
        for (int i = mx; i&&p < tot - 1; i--) {
            int x = 0, y;
            for (int k = 1, tmp; k*i <= mx && p < tot - 1; k++) {
                if (!vis[tmp = k * i])continue;
                y = find(tmp);
                if (!x)x = y; else if (y != x)f[y] = x, ans += i, ++p;
            }
        }
        printf("Case %d: ", k);
        cout << ans << endl;
    }
    //cin >> t;
}
/*
1
3 4 2
1 2 3 1
2 1 1 4
7 8 9 3
1 1 1 1
1 2 3 4
*/
/*

0*/