Poj 1062 expensive dowry difficult AC

The Chinese Question is wrong.

Key Point: two people with a level gap greater than m cannot trade directly or indirectly, and 1 must be the end of the graph, therefore, the levels of all vertices in the shortest path must be in the range of [level [1]-M, level [1] + M]. Therefore, you can enumerate the levels,

For (I = level [1]-m; I <= level [1]; I ++)

{

Shortest PathAlgorithmMake sure that

The level of each vertex in the shortest path is between [I, I + M ].

}

In this way, the bloody AC is always so inspiring.

View code

# Include <stdio. h>
# Include < String . H>
# Include <stdlib. h>
 Const   Int INF = 1000000000 ;
Int Lev [ 110 ], P [ 110 ];
 Int Map [ 110 ] [ 110 ];
 Int Vis [ 110 ];
 Int Dis [ 110 ];
 Int Dijkstra ( Int Cost [] [110 ], Int N, Int X, Int M)
{
 Int I, J, K, minc;
Memset (VIS, 0 , Sizeof (VIS ));
Dis [ 0 ] = 0 ; Vis [ 0 ] = 1 ;
For (I = 1 ; I <n; I ++) dis [I] = cost [ 0 ] [I];
 For (I = 1 ; I <n; I ++)
{
Minc = inf;
 For (J = 0 ; J <n; j ++)
 If (! Vis [J] & dis [J] <minc)
{
Minc = dis [J];
K = J;
}
 If (Minc = inf) Break ;
Vis [k] = 1 ;
 For (J = 0 ; J <n; j ++)
 If (! Vis [J] & dis [J]> dis [k] + cost [k] [J] & lev[ K]> = x & lev[ K] <= X + M)
Dis [J] = dis [k] + cost [k] [J];
}
 Return Dis [ 1 ];
}
Int Main ()
{
 Int M, N;
 Int I, J, K, X, price, ID;

 While (Scanf ( "  % D  " , & M, & N )! = EOF)
{
 For (I = 0 ; I <= 100 ; I ++)
{
Map [I] [I] =0 ;
 For (J = I + 1 ; J <= 100 ; J ++)
{
Map [I] [J] = map [J] [I] = inf;
}
}
 For (I = 1 ; I <= N; I ++)
{
Scanf ( "  % D  " , & P [I], & lev[ I], & X );
 For (J = 1 ; J <= x; j ++)
{
Scanf ( "  % D  " , & ID, & price );
Map [ID] [I] = price;
}
}
 For (I = 1 ; I <= N; I ++)
Map [ 0 ] [I] = P [I];
Lev [0 ] = Lev [ 1 ];
 Int Low = lev [ 1 ]-M;
 Int Ans = inf;
 For (I = low; I <= lev [ 1 ]; I ++)
{
 Int TMP = Dijkstra (MAP, N + 1 , I, m );
 If (TMP <ans)
Ans = TMP;
}
Printf ( "  % D \ n  " , ANS );
}
 Return   0 ;
}