poj 1734 Floyd算求有向图的最小环

题意：旅游公司要开发一条新的路线 ， 要求这是一个总路程尽可能短的环 ， 并且不能只含两个城市 ， 除开起点外 ，不能重复走之前走过的城市 ， 输出这条路线？

Floyd算法求最小环


代码：
//用floyd算法 ， 求有向图的最小环
#include
#include
#include
#include
using namespace std;

#define INF 0xfffffff
#define maxn 110

int grap[maxn][maxn] , n , m;
int dist[maxn][maxn];
int past[maxn][maxn];
int mincircle;
int path[maxn] , k1 ;

void init()
{
    int i ,j;
    for(i = 1; i<= n; i++)
       for(j = 1; j<= n; j++)
          grap[i][j] =INF;
}

void floyd()
{
    mincircle =INF;
    int i ,j;
    for(i = 1; i<= n; i++)
       for(j = 1; j<= n; j++)
       {
          dist[i][j] =grap[i][j];
          past[i][j] =i;
       }

    for(int k =1; k <= n; k++)  //每个点都成为一次中间点 ，和bellman-ford不一样
    {
       for(i = 1; i<= n; i++) //判断是不是最小环
          for(j = 1; j<= n; j++)
          {
             if(i ==j)  continue;
            if(dist[i][j] != INF &&grap[j][k]!=INF&&grap[k][i]!=INF && mincircle >dist[i][j]+grap[j][k]+grap[k][i])
             {
                mincircle =dist[i][j]+grap[j][k]+grap[k][i];
                k1 =0;
                path[k1++] =i;
                path[k1++] =k;
                path[k1++] =j;
               while(past[i][path[k1-1]] != i)
                {
                   path[k1] =past[i][path[k1-1]];
                   k1 +=1;
                }

               
             }
          }

       for(i = 1; i<= n; i++) //Floyd算法
          for(j = 1; j<= n; j++)
          {
             if(i == k ||j == k)  continue;
            if(dist[i][k] != INF && dist[k][j] != INF &&dist[i][k]+dist[k][j]
             {
                dist[i][j] =dist[i][k]+dist[k][j];
                past[i][j] =past[k][j];
             }
          }
    }
}

int main()
{
   while(scanf("%d %d" , &n , &m) != EOF)
    {
      init();
       int i , x ,y , z;
       for(i = 1; i<= m; i++)
       {
          scanf("%d %d%d" , &x , &y , &z);
         if(grap[x][y] > z)
             grap[x][y] =grap[y][x] = z;
       }

      floyd();

       if(mincircle== INF || mincircle < 0)
         cout<<"No solution."<<endl;
       else
       {
          printf("%d", path[0]);
          for(i = 1; i< k1; i++)
             printf(" %d", path[i]);
         cout<<endl;
       }
    }
    return0;
}