連續子向量的最大和及時間度量

Programming Pearls

8.7

#include <stdio.h>
#include <time.h>
#include <stdlib.h>

#define SCALE 3000

int maxnum(int a, int b);

int main(int argc, char const *argv[])
{
  FILE *fp;
  fp = fopen("maximum.in", "r");
  // int x[] = {1,12,-11,10,-65,54,22,-9,21,5,48,5,-8,-2,56,54,-88,-5,2,-8,554,-56,35,-55,555,-65,-545,-23,48,-5,88,-56,16,-8};
  int *x = (int *)malloc(sizeof(int)*(SCALE+1));
  int xi = SCALE,a = 0,num_in = 0;
  while(xi--){
    fscanf(fp, "%d", &x[a++]);
  }


  clock_t start, end;

  // ***Algorithm-1 cube***
  start = clock();

  int max = 0;
  int length = SCALE;
  int i,j,k;
  for (i = 0; i < length; ++i)
  {
    for (j = i; j < length; ++j)
    {
      int sum = 0;
      for (k = i; k <= j; ++k)
      {
        sum += x[k];
      }
      max = maxnum(max, sum);
    }
  }
  
  // long num = 10000000L;
  // while(num--);

  end = clock();

  double times = (double)(end - start)/CLOCKS_PER_SEC;
  double dend = (double)end;

  printf("\n***Algorithm-1 cube***\n");
  printf("end: %f\n", dend);
  printf("Time consuming: %f\n", times);
  printf("%d\n", max);


  // ***Algorithm-2 square***
  start = clock();

  max = 0;

  for (i = 0; i < length; ++i)
  {
    int sum = 0;
    for (j = i; j < length; ++j)
    {
      sum += x[j];
      max = maxnum(max, sum);
    }
  }

  end = clock();

  times = (double)(end - start)/CLOCKS_PER_SEC;
  dend = (double)end;

  printf("\n***Algorithm-2 square***\n");
  printf("end: %f\n", dend);
  printf("Time consuming: %f\n", times);
  printf("%d\n", max);


  // ***Algorithm-3 linear***
  start = clock();

  max = 0;
  int max_end_here = 0;

  for (i = 0; i < length; ++i)
  {
    max_end_here = maxnum(max_end_here + x[i], 0);
    max = maxnum(max, max_end_here);
  }

  end = clock();

  times = (double)(end - start)/CLOCKS_PER_SEC;
  dend = (double)end;

  printf("\n***Algorithm-3 linear***\n");
  printf("end: %f\n", dend);
  printf("Time consuming: %f\n", times);
  printf("%d\n", max);



  free(x);
  x = NULL;
  return 0;
}

int maxnum(int a, int b)
{
  return a > b ? a : b;
}