Codeforces Round #601 (Div. 2) B Fridge Lockers

//题目要求的是每一个点最少要有两条边连接,所以可以先构成一个环。然后再把剩余的最短的边连接起来
#include<iostream>
#include<algorithm>
using namespace std ;
const int N=100010;
int n,m;
struct edge {
    int num;
    int w;
} a[N];
int T;
bool cmp(edge a,edge b) {
    return a.w<b.w;
}
void solve() {
    int ans=0;
    for(int i=1; i<=n; i++) {
        ans+=2*a[i].w;
    }
    ans+=(m-n)*(a[1].w+a[2].w);
    cout<<ans<<endl;
    for(int i=1; i<n; i++) cout<<i<<" "<<i+1<<endl;
    cout<<n<<" "<<"1"<<endl;
    for(int i=1; i<=m-n; i++) cout<<a[1].num<<" "<<a[2].num<<endl;
}
int main() {
    cin>>T;
    while(T--) {
        cin>>n>>m;
        for(int i=1; i<=n; i++) {
            cin>>a[i].w;
            a[i].num=i;
        }
        if(n<=2||m<n) {
            cout<<"-1"<<endl;
            continue;
        }
        sort(a+1,a+1+n,cmp);
        solve();
    }
    return 0;
}

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转载自www.cnblogs.com/QingyuYYYYY/p/11932797.html