Merge pull request #582 from madihamallick/master

Added Knapsack Problem in C

Author: smv1999

Dynamic Programming/Knapsack_greedyMethod.c

@@ -0,0 +1,75 @@
+/*Write program to find the maximum  profit and the solution vector for Knapsack problem. 
+  Consider the following constraints for knapsack. N=3,  M=20
+  Profit : 25 24 15
+  Weight : 18 15 10
+*/
+
+#include<stdio.h>
+#include<conio.h>
+int main(){
+	int p[100],w[100],n,m,i,j,b,s,c;
+	float Iv[100],a,profit,mul[100];
+	printf("Enter knapsack capacity:  ");
+	scanf("%d",&m);
+	printf("\nEnter number of data items:	");
+	scanf("%d",&n);
+	//profit array
+	printf("\nEnter the values of profit array\n");
+	for(i=1;i<=n;i++){
+		printf("Enter p[%d] data item: ",i);
+		scanf("%d",&p[i]);
+	}
+	//weight array
+	printf("\nEnter the values of weight array\n");
+	for(i=1;i<=n;i++){
+		printf("Enter w[%d] data item: ",i);
+		scanf("%d",&w[i]);
+	}
+	//I (profit/weight) value array
+	for(i=1;i<=n;i++){
+		Iv[i]= ((float)p[i]/(float)w[i]);
+	}
+	//sorting Iv,weight, and profit array
+	for (i = 1; i <= n; i++){
+		for (j = 1; j <= n; j++){
+			if (Iv[j] < Iv[i]){
+				a = Iv[i];
+				b = w[i];
+				c = p[i];
+				Iv[i] = Iv[j];
+				w[i] = w[j];
+				p[i] = p[j];
+				Iv[j] = a;
+				w[j] = b;
+				p[j]= c;
+			}
+		}
+	}
+	for(i=1;i<=n;i++){
+		if(m==0){
+			mul[i] = 0;
+			break;
+		}
+		else {
+			s = m - w[i];
+			if (s>0){		
+				m = m - w[i];
+				mul[i] = 1;
+			}
+			else{
+				mul[i] = ((float)m/w[i]);
+				m = m - (w[i] * ((float)m/w[i]));
+			}
+		}
+	}
+	for(i=1;i<=n;i++){
+		profit += (mul[i] * p[i]);
+	}
+	printf("\nHence the total profit will be :  %.2f\n",profit);	
+	//printing array Iv again
+	printf("\nThe solution vector will be \n");
+ 	for(i=1;i<=n;i++){
+		printf("%.2f ",Iv[i]);
+	}
+ 	return 0;
+}