Shortest circuit template (dijkstra&&flody)

1.dijkstra

void Dijkstra (int x,int N)//x represents the starting point, and N represents the total number of {    int min=0,p=0;    for (int i=1;i<=n;i++)    {        dis[i]=map[x][i];        visited[i]=0;    }    Visited[x]=1;    for (int i=1;i<=n;i++)    {        min=inf;        for (int j=1;j<=n;j++)        {            if (!visited[j]&&dis[j]<min)            {                min=dis[j];                p=j;            }        }        Visited[p]=1;        for (int j=1;j<=n;j++)        {            if (!visited[j]&&dis[j]>dis[p]+map[p][j])                Dis[j]=dis[p]+map [P] [j];}}    }

2.flody

void Floyd () {    int i,j,k;    for (k=0;k<n;k++) for (        i=0;i<n;i++) for (            j=0;j<n;j++)                if (Dis[i][j]>dis[i][k]+dis[k][j] )                   dis[i][j]=dis[i][k]+dis[k][j];}

Personal Advice if you can use Dijkstra where possible, flody will usually time out. Of course, if the case of negative ring, you need to use the SPFA algorithm.

Shortest circuit template (dijkstra&&flody)