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AtCoder Beginner Contest 210 E - Ring MST「Kruskal」「加速合并」「同余与gcd」

AtCoder Beginner Contest 210

E - Ring MST

题目描述:

思路:

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
#define int long long
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k, x;
struct ran{
    int a, c;
    bool operator < (const ran &x)const{
        return c < x.c;
    }
}tr[MAX];

int gcd(int a, int b){
    if(!b)return a;
    else return gcd(b, a % b);
}

void work(){
    cin >> n >> m;
    for(int i = 1; i <= m; ++i)cin >> tr[i].a >> tr[i].c;
    sort(tr + 1, tr + 1 + m);
    ll ans = 0;
    int sum = n;
    int gg = n;
    for(int i = 1; i <= m; ++i){
        gg = gcd(gg, tr[i].a);
        ans += (sum - gg) * tr[i].c;
        sum = gg;
        if(gg == 1)break;
    }
    if(gg == 1)cout << ans << endl;
    else cout << -1 << endl;
}

signed main(){
    io;
    work();
    return 0;
}
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