POJ 1695

//和双调欧几里德旅行商一样的做法。参见前面的双调欧几里德旅行商问题
#include < iostream >
#define  REP(i,n) for(int i=0;i<n;i++)
#define  FOR(i,L,R) for(int i=L;i<=R;i++)
#define  INF 10000000
using   namespace  std;
int  cas;
int  dp[ 31 ][ 31 ][ 31 ];
int  map[ 31 ][ 31 ];
int  n;
int  main()
{
    scanf( " %d " , & cas);
     while (cas -- )
    {
        memset(map, 0 , sizeof (map));
        scanf( " %d " , & n);
        REP(i,n)
            REP(j,n)
            {
                 if (j > i)
                    scanf( " %d " , & map[i][j]);
                fill_n(dp[i][j],n,INF);
            }
        dp[ 0 ][ 0 ][ 0 ] = 0 ;
        REP(i,n)
        {
            REP(j,i + 1 )
            {
                REP(k,j + 1 )
                {
                    dp[i + 1 ][j][k] = min(dp[i + 1 ][j][k],dp[i][j][k] + map[i][i + 1 ]); // to i
                    dp[i + 1 ][i][k] = min(dp[i + 1 ][i][k],dp[i][j][k] + map[j][i + 1 ]); // to j
                    dp[i + 1 ][i][j] = min(dp[i + 1 ][i][j],dp[i][j][k] + map[k][i + 1 ]); // to k
                }
            }
        }
         int  ma = INF;
        REP(i,n)
            REP(j,i + 1 )
                ma = min(ma,dp[n - 1 ][i][j]);
        printf( " %d\n " ,ma);
    }
     return   0 ;
}