Jamie and Interesting Graph

看上去很复杂,其实你只要把构造一条链,使得他们的和为1e6+3(>n的质数),就好了,然后其他边为1e9,然后就ac了.
代码如下:

#include <bits/stdc++.h>
using namespace std;
const int N=2e5+5;
struct vv{
    int to,val;
};
vector<vv>v[N];
map<int,int>mp[N];
int main()
{
    int n,m;//n个点 m条边
    scanf("%d%d",&n,&m);
    int ans=1e6+3;
    printf("%d %d\n",ans,ans);
    for(int i=2;i<=n-1;i++)
    {
        v[i-1].push_back({i,1});
        mp[i-1][i]=1;
    }v[n-1].push_back({n,ans-(n-2)});
    mp[n-1][n]=1;//标记x->y有条边.
    m-=(n-1);
    int flag=0;
    for(int i=1;i<=n;i++)
    {
        if(flag) break;
        for(int j=i+1;j<=n;j++)
        {
            if(mp[i][j]) continue;
            if(m==0) {flag=1;break;}
            v[i].push_back({j,(int)1e9});
            mp[i][j]=1;
            m--;
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<v[i].size();j++)
        {
            int k1=v[i][j].to;
            int k2=v[i][j].val;
            printf("%d %d %d\n",i,k1,k2);
        }
    }
    return 0;
}