HDU1874 畅通工程续

简短的最短路问题

  Dijkstra算法求解

#include<iostream>
#include<cstdio>
using namespace std;
const int N=202;
const int INF=1<<28;
int path[N][N],d[N];
bool s[N];
void Dijkstra(int first,int last,int n)
{
   memset(s,false,sizeof(s));
   int i,j,u;
   for(i=0;i<n;i++)
       d[i]=path[first][i];
   d[first]=0;s[first]=true;
   for(i=0;i<n;i++)
   {
      int minn=INF;
      for(j=0;j<n;j++)
          if(!s[j]&&d[j]<minn)
              minn=d[u=j];
          s[u]=true;
          for(j=0;j<n;j++)
              if(!s[j]&&d[j]>d[u]+path[u][j])
                  d[j]=d[u]+path[u][j];
   }

}
int main()
{
    int i,j,n,m,a,b,x,start,end;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                path[i][j]=INF;
       for(i=0;i<m;i++)
       {
         scanf("%d%d%d",&a,&b,&x);
         if(x<path[a][b])
             path[a][b]=path[b][a]=x;
       }
       scanf("%d%d",&start,&end);
       Dijkstra(start,end,n);
       if(d[end]==INF) printf("-1\n");
       else printf("%d\n",d[end]);
    
    }
   return 0;
}

floyd算法求解

#include<iostream>
#include<cstdio>
const int N=202;
const int INF=1<<28;
int path[N][N];
int floyd(int s,int t,int n)
{
   int i,j,k;
   for(k=0;k<n;k++)
       for(i=0;i<n;i++)
       {
           if(path[i][k]!=INF)//优化，减少不必要的时间损耗
            for(j=0;j<n;j++)
                if(path[i][j]>path[i][k]+path[k][j])
                    path[i][j]=path[i][k]+path[k][j];
       }
    return path[s][t];
}
int main()
{
   int i,j,n,m,a,b,x,s,t,ans;
   while(scanf("%d%d",&n,&m)!=EOF)
   {
       for(i=0;i<n;i++)
           for(j=0;j<n;j++)
           {
               if(i==j) path[i][j]=0;
              else  path[i][j]=INF;
           }
      for(i=0;i<m;i++)
      {
           scanf("%d%d%d",&a,&b,&x);
           if(x<path[a][b])
               path[a][b]=path[b][a]=x;
      }
      scanf("%d%d",&s,&t);
      ans=floyd(s,t,n);
      printf("%d\n",ans<INF?ans:-1);
   }
   return 0;
}