HDU 4781/build a graph that meets the requirements

Ah, for the first time, we can see the number of points and the number of edges so that the directed graph can be restored as required.

Requirements:

1. Strong connectivity.

2. There is only one directed edge between any two points, and you and yourself are boundless.

3. Any closed loop right and % 3 are 0.

4. The weight theory of each edge is different, and the value is 1, 2, 3. M.

At the beginning, I thought there would be a big ring 1-> 2-> 3->... N, N-> 1. When I simulated the game, I didn't think about it anymore ..

Add edge first: I-> I + 1 is the right to I, and then select a right to add n to 1 from (n, n + 1, n + 2, make the closed loop % 3 = 0.

Then process the remaining M-n edges,

: Note that if an edge is appended to I ---> J, w (I, j) % 3 = sum of edge weights from I to J % 3 (easy to prove) exists)

Enumerate each entry (n <80, m <n * n/7 ).

#include<iostream>#include<cstring>using namespace std;int n,m;int mark[6400];int vis[85][85];   int main(){    int T;    cin>>T;int ct=1;    while(T--)    {        for(int i=0;i<=m;i++)            mark[i]=0;        memset(vis,0,sizeof(vis));           cin>>n>>m;        for(int i=1;i<n;i++)        {           // cout<<i<<" "<<i+1<<" "<<i<<endl;            mark[i]=1;vis[i][i+1]=vis[i+1][i]=i;        }        int tei=0;        for(int i=n;i<n+3;i++)           if((n*(n-1)/2+i)%3==0)           {              //cout<<n<<" "<<1<<" "<<i<<endl;              tei=i;              mark[i]=1;              vis[n][1]=vis[1][n]=i;              break;          }       for(int i=n;i<=m;i++)       {           if(!mark[i])           {               int flags=0;               for(int j=1;j<=n;j++)               {                for(int k=j+2;k<=n;k++)                {                    if(!vis[j][k]&&((k-j)*(j+k-1)/2)%3==i%3)                    {                        vis[j][k]=vis[k][j]=i;                        mark[i]=1;                        flags=1;                        break;                    }                }                if(flags)break;               }           }       }       bool flag=1;       for(int i=1;i<=m;i++)       {           if(mark[i]==0)             {                 flag=0;break;             }       }       cout<<"Case #"<<ct++<<":"<<endl;       if(flag==0)       cout<<-1<<endl;       else       {           for(int i=1;i<=n;i++)             for(int j=(i+1);j<=n;j++)             {                 if(vis[i][j]!=0&&!(i==1&&j==n))                 {                     cout<<i<<" "<<j<<" "<<vis[i][j]<<endl;                 }             }            cout<<n<<" "<<1<<" "<<tei<<endl;       }    }}

HDU 4781/build a graph that meets the requirements