zadania/samsung/t2/main.c

#include <stdio.h>
#include <math.h>
#include <string.h>

// #define BIG_SIZE (64) // that is 18 446 744 073 709 551 616 paths to check
#define BIG_SIZE (32) // that is 4 294 967 295 paths to check, takes about 30 seconds on i7 3.60GHz

// NOTE: This is probably not the best way to do this.
// A tricky case example is:
// X = 1, Y = 100,
// A = {1,  10, 1,  1,  1},
// B = {10, 1,  1,  1,  1000}

// NOTE: The idea is to start from the end (N-1) and assume that we
// have already found the best paths for previous nodes and pick the
// lower sum for current nodes. While checking previous nodes the 
// function is called recursively so it again assumes that previous 
// nodes are already solved. Since the first pair of nodes are just 
// integers to be compared, we stop the recursion and the "unwinded"
// stack gives us the answer.

int solution_(int A[], int B[], int N, int X, int Y, char onA)
{
  if (N == 0) {
    return onA ? A[0] : B[0];
  }

  int c1, c2;
  if (onA) {
    c1 = 0 + A[N] + solution_(A, B, N-1, X, Y, 1);
    c2 = Y + A[N] + solution_(A, B, N-1, X, Y, 0);
  }
  else {
    c1 = 0 + B[N] + solution_(A, B, N-1, X, Y, 0);
    c2 = X + B[N] + solution_(A, B, N-1, X, Y, 1);
  }

  return (c1 <= c2 ? c1 : c2);
}

int solution(int A[], int B[], int N, int X, int Y)
{
  int sa = solution_(A, B, N-1, X, Y, 1);
  int sb = solution_(A, B, N-1, X, Y, 0);
  return (sa <= sb ? sa : sb);
}

void test(int actual, int expected)
{
  if (actual != expected) {
    printf("Test failed, expected %i, got %i\n", expected, actual);
  }
  else {
    printf("Test passed (expected %i, got %i)\n", expected, actual);
  }
}

int main()
{
  {
    int A[] = {1, 6};
    int B[] = {3, 2};
    int N = 2, X = 2, Y = 10;
    test(solution(A, B, N, X, Y), 5);
  }

  {
    int A[] = {1, 6, 2};
    int B[] = {3, 2, 5};
    int N = 3, X = 2, Y = 1;
    test(solution(A, B, N, X, Y), 8);
  }

  {
    int A[] = {2, 11, 4, 4};
    int B[] = {9, 2, 5, 11};
    int N = 4, X = 8, Y = 4;
    test(solution(A, B, N, X, Y), 21);
  }

  {
    int A[] = {1, 10, 1};
    int B[] = {10, 1, 10};
    int N = 3, X = 1, Y = 5;
    test(solution(A, B, N, X, Y), 9);
  }

  {
    int A[] = {8, 3, 3};
    int B[] = {6, 1, 10};
    int N = 3, X = 4, Y = 3;
    test(solution(A, B, N, X, Y), 13);
  }

  {
    int A[] = {1, 10, 1, 1};
    int B[] = {10, 1, 1, 100};
    int N = 4, X = 1, Y = 100;
    test(solution(A, B, N, X, Y), 13);
  }

  {
    int A[BIG_SIZE];
    int B[BIG_SIZE];
    int N = BIG_SIZE, X = 1, Y = 100;
    memset(A, 0, sizeof(A));
    memset(B, 0, sizeof(A));
    A[0] = 1;
    B[0] = 10;
    B[BIG_SIZE-1] = 100;
    test(solution(A, B, N, X, Y), 1);
  }


  printf("Done\n");
}