UV 11997 K smallest sums

Uva_11997

Because K-row processing is cumbersome, so a feasible idea is to treat as two rows, so that the processing of K-1 times can be.

When there are only two rows, you can consider how to construct the sum one by one from small to large. This can be achieved through the priority queue: first put the two smallest values in, then, the minimum and the two numbers are taken out each time, and the first number is changed to a slightly larger number in the first row, and then placed in the priority queue, then try to replace the second number with a larger number in the second row into the priority queue. In this way, the first k elements retrieved from the priority queue are the sum of the first K.

# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Define Mind 562510 # Define Minks 760 # Define INF 0x3f3f3f Int N, D, Min [ 4 * Mind], a [mink] [mink], ANS [mink];  Void  Init (){  Int  I, J;  For (I = 0 ; I <n; I ++ ){  For (J = 0 ; J <n; j ++) scanf ( "  % D " ,& A [I] [J]); STD: Sort (A [I], a [I] + N );}  For (D = 1 ; D <n * n + 2 ; D <= 1  ); Memset (Min,  0x3f , Sizeof (Min [ 0 ]) * 2 * D );} Void Update ( Int  I ){  For (; I ^ 1 ; I >>= 1  ) Min [I > 1 ] = STD: min (Min [I], Min [I ^ 1  ]);}  Int  GETID (){  Int  I;  For (I = 1 ; I < D ;){  If (Min [I < 1 ] = Min [I]) I <= 1  ;  Else I = I < 1 | 1  ;}  Return  I ;}  Void  Pop (){ Int I = GETID (); Min [I] = INF, update (I );}  Void Push ( Int X, Int Y, Int  V ){  Int I = x * n + Y; Min [d + I] = V, update (D + I );}  Void Top ( Int & X,Int & Y, Int & V ){  Int I = GETID ()- D; x = I/N, y = I % N, V = min [d + I];}  Void  Solve (){  Int  I, j, X, Y, V, CNT;  For (I = 1 ; I <n; I ++ ){  While (Min [ 1 ]! = INF) Pop (); CNT = 0  ; Push (  0 , 0 , A [I- 1 ] [ 0 ] + A [I] [ 0  ]);  While (CNT < N) {top (X, Y, v), pop (); ans [CNT ++] = V; If (X <n) Push (x + 1 , Y, a [I- 1 ] [X + 1 ] + A [I] [Y]);  If (Y <n) Push (X, Y + 1 , A [I- 1 ] [X] + A [I] [Y + 1  ]);}  For (J = 0 ; J <n; j ++) A [I] [J] = Ans [J];} printf ( "  % D  " , A [n- 1 ] [ 0  ]);  For (I = 1 ; I <n; I ++) printf ( "  % D  " , A [n- 1  ] [I]); printf (  "  \ N  " );}  Int  Main (){  While (Scanf ( "  % D  " , & N) = 1  ) {Init (); solve ();}  Return   0  ;}