C Implementation of adversarial search ( MIN-MAX )

adversarial.c

/**
 * Implementation of adversarial search
 * Example To Use Program :https://s15.postimg.org/yyh4lhdsr/Adversial.png
 *
 * Email: engineer.rc1@gmail.com 
 * @author Tomrock D'souza, St. Francis Institute Of Technology, University of Mumbai, 2017
 * @copyright GNU General Public License v3.0
 * No reproduction in whole or part without maintaining this copyright notice
 * and imposing this condition on any subsequent users.
 */
#include<stdio.h>
#include<conio.h>
int poop[20], lol[20][20], p = 0;

//This Function Creates the tree.
void admat() {
  int i, j, k;
  for (i = 0; i <= p; i++) {
    printf("Enter number of daughter nodes of node%d: ", i);
    scanf("%d", & j);
    if (j != 0) {
      printf("Nodes Assigned to node %d: ", i);
      for (k = 0; k < j; k++) {
        p++;
        lol[i][p] = 1;
        printf("%d ", p);
      }
      printf("\n");
    }
  }
}

//This Function Decides the Highest or lowest From the Sub-Branches based on the Level
int minmax(int a, int b, int s) {
  int d;
  if (s == 0) {
    if (a < b) {
      d = a;
    } else {
      d = b;
    }
  } else if (s == 1) {
    if (a < b) {
      d = b;
    } else {
      d = a;
    }
  } else d = a;

  return d;
}

//This Function parses the tree and askes the user value of a node if it has no children
int trav(int n, int a, int s) {
  int i, b = 0, c = 0, e = 0;
  s = (s + 1) % 2;
  //This loop iterates through all child nodes 1 by 1
  for (i = 0; i < n; i++) {
    if (lol[a][i] == 1) {
      e = 1;
      if (b == 0) {
		  //This Value is assigned 2 because it's a single node
        c = minmax(trav(n, i, s), c, 2);
        b = 1;
      } else {
        c = minmax(trav(n, i, s), c, s);
      }
    }
  }
  if (e == 0) {
    printf("Enter value of node %d:", a);
    scanf("%d", & c);
    poop[a] = c;
    return c;
  } else {
    poop[a] = c;
    return c;
  }
}

int main() {
  int i, j;
  admat();
  trav(p+1, 0, 0);
  //Print The Values of The Nodes After The adversarial search is Complete
  for (i = 0; i <=p; i++) {
    
  }
  printf("\nOUTPUT\n-------------------------\n");
  printf("Value of Root Node0 is %d\n\n",poop[0]);
   for (i = 0; i <=p; i++) {
	   for (j = 0; j <=p; j++) {
		   if(lol[i][j]){
			   printf("Node%d => Node%d Value %d\n", i,j, poop[j]);
		   }
	   }
	   printf("\n");
  }
  getch();
  return 0;
}